ECO101 Final Exam Prep

Week 1

Lecture

ECO101 - Week 01 Equilibrium.pdf

ECO101 Lecture 01
1
2
- $D = 10 - 2 p_{x} - 3 p_{y} + 5 m$
- When $p_{y} + +, D - -$
- When $p_{x} + +, D - -$
- Good y is a complement in consumption for good x
- Examples of complements in production:
  - Beef and leather
- Example of substitute in production:
  - Sedans and SUVS
3
- $D_{x} = 10 - 2 p_{x} + 3 p_{y} - 5 m$
- $p_{x} ↑, D_{x} ↓$
- $p_{y} ↑, D_{x} ↑$
- See that good Y is a substitute in consumption for good X
4
5
- Consumer with $m = 24$
- Purchase blueberries $Q_{b} = 2$ with $p_{b} = 3$
- and strawberries with $Q_{s} = 3$ at $p_{s} = 6$
- Relative price of strawberries is?
- $\frac{6}{3} = 2$
- It's $2$
6
7
9
- An increase in supply can be caused by any of the following except:
  - A rise in the price of a complement in production
  - A fall in the price of a substitute in production
  - An improvement in technology
  - An increase in the price of an input to production
11
- $D = 12 - 2 p$
- $S = 2 p - 4$
- $D = S$
- $12 - 2 p = 2 p - 4$
- $p = 4$
- $2 (4) - 4 = 4$
- $(4, 4)$
12
- $D = 12 - 2 p$
- $S = 2 p - 4$
- $p_{0} = 5$
- $p_{1} = 3$
- $D (p_{0}) = 12 - 2 (5) = 2$
- $S (p_{0}) = 2 (5) - 4 = 6$
- $D (p_{1}) = 12 - 2 (3) = 6$
- $S (p_{1}) = 2 (3) - 4 = 2$
- $(2, 6)$
- $(6, 2)$
13
- $D_{x} = 16 - 2 p_{x} + 4 p_{y}$
- $S_{x} = 2 p_{x} - 2 w$
- $p_{y_{0}} = 1$
- $p_{y_{1}} = 3$
- $w = 2$
- $p_{y_{0}} = 1$
  - $D_{x} (p_{y_{0}}) = 16 - 2 p_{x} + (4) (1) = 20 - 2 p_{x}$
  - $S_{x} (p_{y_{0}}) = 2 p_{x} - (2) (2) = 2 p_{x} - 4$
  - $D_{x} = S_{x}$
  - $20 - 2 p_{x} = 2 p_{x} - 4$
  - $p_{x} = 6$
  - $D (6) = 20 - 2 (6) = 8$
  - $(6, 8)$
- $p_{y_{1}} = 3$
  - $D_{x} (p_{y_{1}}) = 16 - 2 p_{x} + (4) (3) = 28 - 2 p_{x}$
  - $S_{x} (p_{y_{1}}) = 2 p_{x} - 2 (2) = 2 p_{x} - 4$
  - $D_{x} = S_{x}$
  - $28 - 2 p_{x} = 2 p_{x} - 4$
  - $p_{x} = 8$
  - $D_{x} (8) = 28 - 2 (8) = 12$
  - $(8, 12)$
```
	left=-0.5; right=18;
	top=30; bottom=-0.5;
	---
	y=28-2x
	y=20-2x
	y=2x-4
	(6,8)
	(8,12)
```
15
- An economist will infer that a higher Equilibrium price and lower quantity could be caused by: an increase in the price of substitute in production.

Tutorial

1
- $S = 0.5 p - 20$
- $Q_{S} = 0.5 p - 20$
- $x = 0.5 y - 20$
- $x + 20 = 0.5 y$
- $2 x + 40 = y$
- $D = 80 - 0.5 p + m$
- $m_{0}$
  - $x = 80 - 0.5 y + 20$
  - $y = 200.0 - 2 x$
- $m_{1}$
  - $x = 80 - 0.5 y + 40$
  - $y = 240.0 - 2 x$
- $m_{0} = 20$
  - $S = D$
  - $0.5 p - 20 = 80 - 0.5 p + 20$
  - $p = 120.0$
  - $S (120) = 0.5 (120) - 20 = 40.0$
- $m_{1} = 40$
  - $S = D$
  - $0.5 p - 20 = 80 - 0.5 p + 40$
  - $p = 140.0$
  - $S (140) = 0.5 (140) - 20 = 50.0$
```
    left=-0.5; right=180;
    top=250; bottom=-0.5;
    ---
    y=240-2x
    y=200-2x
    y=2x+40
    (50,140)
    (40,120)
```

Applications

Week 2

Lecture

ECO101 - Week 02 Elasticity.pdf
ECO101 Lecture 02

1
- 3 factors that lead to an inelastic demand curve
  - If you require this thing
  - If there's no substitutes
  - Lag
- How it's related to slope
  - If you're inelastic, it means your slope is higher.
2
- $D_{x} = 16 - 2 p_{x} + 4 p_{y}$
- $p_{y} = 1$
- $S_{x} = 2 p_{x} - 2 w$
- $w = 2$
- Find elasticity from $p = 4 \to 6$
- $D_{x} = 20 - 2 p_{x}$
- $S_{x} = 2 p_{x} - 4$
- $p_{0}$
  - $D_{x} = 20 - (2) (4) = 12$
  - $S_{x} = (2) (4) - 4 = 4$
- $p_{1}$
  - $D_{x} = 20 - (2) (6) = 8$
  - $S_{x} = (2) (6) - 4 = 8$
- $η = \frac{\frac{Δ Q}{\bar{Q}}}{\frac{Δ p}{\bar{p}}}$
- $η = \frac{\frac{12 - 8}{(12 + 8) (0.5)}}{\frac{(4 - 6)}{(0.5) (4 + 6)}} = - 1.0 = 1$
5
- $D = 240 - 20 p$
- $S = 20 p$
- Find ownprice elasticity of demand
- $p = 2 \to 6$
- $p_{0}$
  - $D = 240 - 20 (2) = 200$
- $p_{1}$
  - $D = 240 - 20 (6) = 120$
- $η = \frac{\frac{(200 - 120)}{(0.5) (200 + 120)}}{\frac{(6 - 2)}{(0.5) (6 + 2)}} = 0.5$
6
- When we have unit elasticity, we have max revenue.
8
- $D = 160 - 2 p$
- Total expenditures on this good are at a maximum when sellers charge a price of $x$
- We know when we have unit elasticity, we have max spend.
- $η = \frac{\frac{Δ Q}{\bar{Q}}}{\frac{δ p}{\bar{p}}}$
- On a linear demand curve, this is the mid point
- $x = 160 - 2 y$
- $x - 160 = - 2 y$
- $- \frac{1}{2} x + 80 = y$
- $\frac{1}{2} x = 80$
- $x = 160$
- $(160, 0)$
- So $\frac{160}{2} = 80$
- $80 = 160 - 2 p$
- $p = 40$ is when we maximize spend.

Tutorial

1
- $D = 20 - 2 p$
- Find price elasticity from $p = 1 \to 3$
- $D (1) = 20 - 2 (1) = 18$
- $D (3) = 20 - 2 (3) = 14$
- $η = \frac{\frac{Δ Q}{\bar{Q}}}{\frac{Δ p}{\bar{p}}}$
- $η = \frac{\frac{(18 - 14)}{(0.5) (18 + 14)}}{\frac{(3 - 1)}{(0.5) (3 - 1)}} = 0.125$
2
- $D = 80 - 2 p$
- $S = 2 p - 20$
- Calculate the price at which demand is unit elastic
- $x = 80 - 2 y$
- $y = 40 - \frac{1}{2} x$
- $40 = \frac{1}{2} x$
- $x = 80$
- $40 = 80 - 20 p$
- $p = 2$
- At $2 $$ it's unit elastic.
3
- $D = 80 - p$
- $S = p - 20$
- Find max expenditure
- When it's unit elastic
- $x = 80 - y$
- $y = - x + 80$
- $x = 80$
- $40 = 80 - p$
- $p = 40$
- $(40, 40)$ is the point
4
- $D_{x} = 5 - p_{x} + p_{y}$
- $S_{x} = p_{x}$
- $p_{y} = 1 \to 3$
- Find the cross-price elasticity for good $x$
- $p_{y_{0}}$
  - $D_{x} = 5 - p_{x} + 1 = 6 - p_{x}$
  - $S_{x} = p_{x}$
  - $6 - p_{x} = p_{x}$
  - $p_{x} = 3$
  - $D_{x} (3) = 6 - 3 = 3$
  - $(3, 3)$
- $p_{y_{1}}$
  - $D_{x} = 5 - p_{x} + 3 = 8 - p_{x}$
  - $S_{x} = p_{x}$
  - $D_{x} = 8 - (3) = 5$
- $η_{x y}^{D} = \frac{\frac{Δ Q_{x}}{{\bar{Q}}_{x}}}{\frac{Δ p_{y}}{{\bar{p}}_{y}}}$
- $η_{x y}^{D} = \frac{\frac{(5 - 3)}{(0.5) (5 + 3)}}{\frac{(3 - 1)}{(0.5) (3 + 1)}} = 0.5$

Applications

Week 3

ECO101 - Week 03 Regulation.pdf

Lecture

2
- $D = 12 - 2 p$
- $S = 2 p$
- $t = 2$
- a
  - $D = S$
  - $12 - 2 p = 2 p$
  - $p = 3$
  - $D (3) = 12 - 2 (3) = 6$
  - $(6, 3)$
- b
  - After tax
  - $D = S - t$
  - $12 - 2 p = (2) (p - t)$
  - $p = \frac{t}{2} + 3$
  - $p = \frac{2}{2} + 3 = 4$
  - $D (4) = 12 - 2 (4) = 4$
  - $(4, 4)$
  - $\begin{array}{l} p^{*} & t = 2 & Δ \\ CS \\ PS \\ GS \\ TS \end{array}$
  - $C S$
    - Regular
      - $(6 - 0) (3 - 0) (0.5) = 9$
    - Taxed
      - $(4 - 0) (6 - 4) (0.5) = 4$
  - $P S$
    - Regular
      - $9$
    - Taxed
      - $4$
  - $G S$
    - Regular
      - $0$
    - Taxed
      - $b h$
      - $(4 - 3) (4 - 0) = 4$
  - $T S$
    - $9 + 9 + 0 = 18$
    - $4 + 4 + 4 = 12$
```
    left=-0.5; right=15;
    top=15; bottom=-0.5;
    ---
    x=12-2y
    x=2y
    x=2(y-2)
    (6,3)
    (4,4)
```
3
- $D = 12 - 2 p$
- $S = 2 p$
- $s = 2$
- a
  - Before Subsidy
    - $D = S$
    - $12 - 2 p = 2 p$
    - $p = 3$
    - $D (3) = 12 - 2 (3) = 6$
    - $(6, 3)$
  - After Subsidy
    - $D = S + s$
    - $12 - 2 p = (2) (p + s)$
    - $p = 3 - \frac{s}{2} |_{s = 2}$
    - $p = 3 - 1 = 2$
    - $D (2) = 12 - 2 (2) = 8$
    - $(8, 2)$
- b
  - Consumer Surplus
    - $s = 0$
      - $\frac{b h}{2} = \frac{(6 - 0) (6 - 3)}{2} = 9$
    - $s = 2$
      - $\frac{b h}{2} = \frac{(8 - 0) (6 - 2)}{2} = 16$
  - Producer Surplus
    - $s = 0$
      - $\frac{b h}{2} = \frac{(3 - 0) (6 - 0)}{2} = 9$
    - $s = 2$
      - $\frac{b_{0} h_{0}}{2} + b_{1} h_{1} = \frac{(8 - 4) (2 - 0)}{2} + (4 - 0) (2 - 0) = 12$
      - $\frac{b h}{2} = 16$
  - Government Surplus
    - $s \cdot q_{s = 2}$
    - $2 (8) = - 16$
  - Total Surplus
    - $9 + 9 + 0 = 18$
    - $16 + 12 - 16 = 12$
    - $16 + 16 - 16 = 16$
```
    left=-0.5; right=15;
    top=8; bottom=-0.5;
    ---
    x=12-2y
    x=2y
    x=2(y+2)
    (6,3)
    (8,2)
    (0,6)
    (0,3)
    (0,2)
    y=3|dashed|0<=x<=6
    y=2|dashed|0<=x<=8
```

Tutorial

1
- $D = 40 - p$
- $S = p$
- Calculate consumer surplus when a price ceiling is introduced at $p = 10$
- Pre-ceiling
  - $D = S$
  - $40 - p = p$
  - $p = 20 ⟹ Q_{D} = 20$
  - $(20, 20)$
- Post ceiling
  - $D (10) = 40 - 10 = 30$
    - $(30, 10)$
  - $S (10) = 10$
    - $(10, 10)$
- Consumer Surplus
  - Pre-ceiling
    - $\frac{b h}{2}$
    - $\frac{(20 - 0) (40 - 20)}{2} = 200$
  - Post-ceiling
    - $\frac{b h}{2} + b h$
    - $\frac{(40 - 30) (10 - 0)}{2} + (10 - 0) (30 - 10) = 250$
- Efficiency Loss
  - Pre = $0$
  - Post
    - $\frac{b h}{2} = \frac{(20 - 10) (30 - 10)}{2} = 100$
```
    left=-0.5; right=45;
    top=45; bottom=-0.5;
    ---
    f(y)=40-y
    g(y)=y
    (20,20)
    y=10|dotted
    (10,10)
    (30,10)
    x=10|dashed|0<=y<=10
    x=30|dashed|0<=y<=10
    x<f(y)|x<g(y)|y>20
    x<f(y)|x<g(y)|y>10
	y < 40 - x \{ y > x \} \{ x > 10 \}
```

$S = 0.5 w - 20$
$D = 280 - 0.5 w$
$w \geq 400$
Initial
- $D = S$
- $280 - 0.5 w = 0.5 w - 20$
- $w = 300.0 ⟹ q = 130$
Post
- $D (400) = 280 - 0.5 (400) = 80.0$
- $S (400) = 0.5 (400) - 20 = 180.0$
Points
- $(130, 300)$
- $(180, 400)$
- $(80, 400)$

Calculate efficiency loss

$\frac{b h}{2} = \frac{(130 - 80) (400 - 200)}{2} = 5000$

    left=-0.5; right=300;
    top=600; bottom=-0.5;
    ---
    f(y)=0.5y-20
    g(y)=280-0.5y
    (130,300)
    (180,400)
    (80,400)
    y=400|dotted
    x=80|dashed|0<=y<=400
    x=180|dashed|0<=y<=400
    x<=g(y)\{x<=f(y)\} \{x>=80\}

3
- $D = 140 - 0.5 p$
- $S = 0.5 p - 20$
- $t = 20$
- Initial
  - $D = S$
  - $140 - 0.5 p = 0.5 p - 20$
  - $p = 160.0 ⟹ q = 60$
  - $(60, 160)$
- After tax
  - $D = S - t$
  - $140 - 0.5 p = 0.5 (p - t) - 20$
  - $p = 0.5 t + 160.0 |_{t = 20}$
  - $p = 0.5 (20) + 160.0 = 170.0 ⟹ q = 55$
  - $(55, 170)$
- Solve for bottom of government revenue
  - $0.5 p - 20 = 55$
  - $p = 150.0$
- Efficiency Loss
  - $\frac{b h}{2} = \frac{(60 - 55) (170 - 150)}{2} = 50$
```
    left=-0.5; right=100;
    top=200; bottom=100;
    ---
    f(y)=140-0.5y
    g(y)=0.5y-20
    h(y)=0.5(y-20)-20
    (60,160)
    (55,170)
    x<=f(y)|x<=g(y)|x>=55
    y=150|dashed|0<=x<=55
    y=170|dashed|0<=x<=55
    x=55|dashed|150<=y<=170
```
4
- $p \leq 20$
- $D = 100 - p$
- $S = p$
- Find efficiency gain if the ceiling was eliminated
- Initial with ceiling
  - $D (20) = 100 - 20 = 80$
  - $S (20) = 20$
  - $(80, 20)$
  - $(20, 20)$
  - CS
    - $\frac{b h}{2} + b h$
    - $\frac{(20 - 0) (100 - 80)}{2} + (20 - 0) (80 - 20) = 1 400$
  - PS
    - $\frac{b h}{2} = \frac{(20 - 0) (20 - 0)}{2} = 200$
  - TS
    - $1 400 + 200 = 1 600$
- Without ceiling
  - $D = S$
  - $100 - p = p$
  - $p = 50 ⟹ q = 50$
  - CS
    - $\frac{b h}{2} = \frac{(50 - 0) (100 - 50)}{2} = 1 250$
  - PS
    - $\frac{b h}{2} = 1 250$
  - TS
    - $2 (1 250) = 2 500$
- $T S = 1600 \to 2500$ illustrating a $56.25 %$ gain in efficiency.

    left=-2; right=105;
    top=105; bottom=-2;
    ---
    f(y)=100-y
    g(y)=y
    (80,20)
    (20,20)
    (50,50)
    y=20|0<=x<=80|dashed
    x=20|0<=y<=20|dashed
    x=80|0<=y<=20|dashed

Applications

Week 4

Lecture

ECO101 - Week 04 International.pdf

Imported good

$D = 12 - 2 p$
$S = 2 p$
$p_{w} = 1$
Equilibrium
- $D = S$
- $12 - 2 p = 2 p$
- $p = 3 ⟹ q = 6$
- $(6, 3)$

See that it's cheaper elsewhere

p_{w} = 1 < p = 3

$D (1) = 12 - 2 (1) = 10$
$S (1) = 2 (1) = 2$
$(10, 1)$
$(2, 1)$

    left=-0.5; right=13;
    top=7; bottom=-0.5;
    ---
    x=12-2y
    x=2y
    (6,3)
    (10,1)
    (2,1)
    y=1|dashed|0<=x<=10
    x=2|dashed|0<=y<=1
    x=10|dashed|0<=y<=1
    y=3|dashed|0<=x<=6

Exported good

$D = 12 - 2 p$
$S = 2 p$
$p_{w} = 5$
Equilibrium
- $D = S$
- $12 - 2 p = 2 p$
- $p = 3 ⟹ q = 6$
- $(6, 3)$
See that it's expensive elsewhere $p_{w} = 5 > p = 3$
- $D (5) = 12 - 2 (5) = 2$
- $S (5) = 2 (5) = 10$
- $(2, 5)$
- $(10, 5)$

Efficiency Gain

$\frac{b h}{2} = \frac{(10 - 2) (5 - 3)}{2} = 8$

    left=-0.5; right=13;
    top=7; bottom=-0.5;
    ---
    x=12-2y
    x=2y
    (6,3)
    (2,5)
    (10,5)
    y=5|dashed|0<=x<=10
    x=2|dashed|0<=y<=5
    x=10|dashed|0<=y<=5
    y=3|dashed|0<=x<=6

Two Country Equilibrium
- $D_{A} = 18 - p$
- $S_{A} = p - 6$
- $D_{B} = 12 - p$
- $S_{B} = p - 4$
- $A$
  - $D_{A} = S_{A}$
  - $18 - p = p - 6$
  - $p = 12 ⟹ q = 6$
  - $(6, 12)$
- $B$
  - $D_{B} = S_{B}$
  - $12 - p = p - 4$
  - $p = 8 ⟹ q = 4$
  - $(4, 8)$
- It's cheaper in country $B$
Import Tariffs
- $D = 12 - 2 p$
- $S = 2 p$
- $p_{w} = 1$
- $t = 1$ tariff
- Initial
  - $D = S$
  - $12 - 2 p = 2 p$
  - $p = 3 ⟹ q = 6$
  - Cheaper elsewhere so we import
  - $(6, 3)$
- Imported
  - $D (1) = 12 - 2 (1) = 10$
  - $S (1) = 2 (1) = 2$
  - $(10, 1)$
  - $(2, 1)$
- Tariffed
  - $D = S - t$
  - $12 - 2 p = (2) (p - t)$
  - $p = \frac{t}{2} + 3$
  - $p = \frac{1}{2} + 3 = \frac{7}{2} = 3.5 ⟹ q = 5$
  - $(5, 3.5)$
- Imported after tariff
  - $D (1) = 12 - 2 (1) = 10$
  - $S (1) = (2) (1 - 1) = 0$
  - $(0, 1)$
  - $(10, 1)$
  - See that after tariff, there's no willingness to supply.
  - So tariff worked to keep people buying domestically.
- Solve for bottom point
  - $5 = 2 p$
  - $p = \frac{5}{2} ⟹ q = 5$
- Consumer Surplus:
  - Pre-tariff
    - $\frac{b h}{2} = \frac{(6 - 3) (6 - 0)}{2} = 9$
  - After tariff
    - $\frac{b h}{2} + b h = \frac{(6 - 5) (2 - 1)}{2} + (2 - 0) (5 - 1) = \frac{17}{2} = 8.5$
- Producer Surplus:
  - Pre-tariff
    - $\frac{b h}{2} = \frac{(6 - 0) (3 - 0)}{2} = 9$
  - After tariff
    - $\frac{b h}{2} = \frac{(2 - 0) (1 - 0)}{2} = 1$
- Government Surplus:
  - Pre-tariff $= 0$
  - After tariff: $b h = (5 - 0) (3.5 - 2.5) = 5$
- Total Surplus:
  - Pre-tariff $= 9 + 9 + 0 = 18$
  - After tariff $= 8.5 + 1 + 5 = 14.5$
```
left=-1; right=13;
top=7; bottom=-1;
---
x=12-2y|Label: Demand
x=2y|Label: Supply
y=1|dashed|Label: World Price (Pw)
y=2|dashed|Label: Pw + Tariff
(6,3)|Label: Autarky
(2,1)|Label: Qs Free Trade
(10,1)|Label: Qd Free Trade
(4,2)|Label: Qs Tariff
(8,2)|Label: Qd Tariff
```
Import Quotas
- $D = 12 - 2 p$
- $S = 2 p$
- $p_{w} = 1$
- Autarky
  - $D = S$
  - $12 - 2 p = 2 p$
  - $p = 3 ⟹ q = 6$
  - $(6, 3)$
- Imported
  - $D (1) = 12 - 2 (1) = 10$
  - $S (1) = 2 (1) = 2$
  - $(10, 1)$
  - $(2, 1)$
- Quota of $2$
  - Max imported can be $2$
  - Area under world price and points is what we want
  - $D (p) - S (p) = 2$
  - $12 - 2 p - 2 p = 2$
  - $p = \frac{5}{2} ⟹ q = 5$
  - $D (\frac{5}{2}) = 12 - (2) (\frac{5}{2}) = 7$
  - $S (\frac{5}{2}) = (2) (\frac{5}{2}) = 5$
  - $(7, \frac{5}{2})$
  - $(5, \frac{5}{2})$
- Consumer Surplus
  - Pre: $\frac{b h}{2} = \frac{(6 - 1) (10 - 0)}{2} = 25$
  - Post: $\frac{b h}{2} = \frac{(6 - 2.5) (7 - 0)}{2} = 12.25$
- Producer Surplus
  - Pre: $\frac{b h}{2} = \frac{(2 - 0) (1 - 0)}{2} = 1$
  - Post: $\frac{b h}{2} = \frac{(5 - 0) (2.5 - 0)}{2} = 6.25$
- Licensee Surplus
  - $b h = (7 - 5) (2.5 - 1) = 3$
- Total Surplus
  - $25 + 1 + 0 = 26$
  - $12.25 + 6.25 + 3 = 21.5$
```
left=-1; right=13;
top=7; bottom=-1;
---
x=12-2y|Label: Demand
x=2y|Label: Supply
y=1|dashed|Label: World Price (Pw)
(6,3)|Label: Autarky
(2,1)|Label: Qs Free Trade
(10,1)|Label: Qd Free Trade
(7,5/2)|Label: Qd Quota
(5,5/2)|Label: Qs Quota
x=5|dashed|0<=y<=5/2
x=7|dashed|0<=y<=5/2
y=5/2|dashed|0<=x<=7
```

Tutorial

1
- $S = p - 2$
- $D = 24 - p$
- $p_{w} = 5$
- $t = 3$ tariff
- Autarky
  - $D = S$
  - $p - 2 = 24 - p$
  - $p = 13 ⟹ q = 11$
  - $(11, 13)$
- Free Trade
  - $D (5) = 24 - 5 = 19$
  - $S (5) = 5 - 2 = 3$
  - $(19, 5)$
  - $(3, 5)$
- Tariff
  - $p_{w} + t |_{t = 3}$
  - $5 + 3 = 8$
  - $D (8) = 24 - 8 = 16$
  - $S (8) = 8 - 2 = 6$
  - $(16, 8)$
  - $(6, 8)$
- $(11, 13)$
- $(19, 5)$
- $(3, 5)$
- $(16, 8)$
- $(6, 8)$
- Efficiency Loss
  - $b h = (6 - 3) (8 - 5) = 9$
  - We can do this because it's symmetrical
- Efficiency loss will depend on the elasticity, when both are very inelastic, we will have the least efficiency loss. When both are very elastic, we have a huge efficiency loss.
```
    left=-2; right=25;
    top=25; bottom=-2;
    ---
	y=4(x-3)+5|dotted
	y=-4(x-19)+5|dotted
    x=y-2
    x=24-y
    (11,13)
    (19,5)
    (3,5)
    (16,8)
    (6,8)
    y=5|dashed|0<=x
    y=8|dashed|0<=x
    x=19|dashed|0<=y<=5
    x=16|dashed|0<=y<=8
    x=3|dashed|0<=y<=5
    x=6|dashed|0<=y<=8
```
2
- $p_{w} = 60$
- $D = 80 - p$
- $S = p$
- Autarky
  - $D = S$
  - $80 - p = p$
  - $p = 40 ⟹ q = 40$
  - $(40, 40)$
- Free trade
  - $p_{w} = 60$
  - $D (60) = 80 - 60 = 20$
  - $S (60) = 60$
  - $(60, 60)$
  - $(20, 60)$
- Points
  - $(40, 40)$
  - $(60, 60)$
  - $(20, 60)$
- Efficiency Gain
  - $\frac{b h}{2} = \frac{(60 - 20) (60 - 40)}{2} = 400$
- This depends on elasticity
  - If both are very elastic, then we have a bigger efficiency gain
```
	left=-2; right=100;
	top=100; bottom=-2;
	---
	x=80-y
	x=y
	y=60|dashed
	y=-4(x-20)+60|dotted
	y=4(x-60)+60|dotted
	(40,40)
	(60,60)
	(20,60)
```
3
- $p_{w} = 10$
- $D = 60 - p$
- $S = p$
- Find consumer surplus
- Autarky
  - $60 - p = p$
  - $p = 30 ⟹ q = 30$
- Free trade
  - $p_{w} = 10$
  - $D (10) = 60 - 10 = 50$
  - $S (10) = 10$
- Consumer Surplus
  - $\frac{b h}{2} = \frac{(50 - 0) (60 - 10)}{2} = 1250$
- Points
  - $(30, 30)$
  - $(50, 10)$
  - $(10, 10)$
- What will give the largest efficiency gain?
  - Supply and demand be very inelastic
```
    left=-2; right=65;
    top=65; bottom=-2;
    ---
    (30,30)
    (50,10)
    (10,10)
    x=60-y
    y=x
	y=4(x-10)+10
	y=-4(x-50)+10
    y=10|dashed
```
4
- $p_{w} = 2$
- $D = 16 - p$
- $S = p$
- Autarky
  - $(8, 8)$
- $q = 6$ quota
  - $D (p) - S (p) = 6$
  - $16 - p - p = 6$
  - $p = 5$
  - $D (5) = 16 - 5 = 11$
  - $S (5) = 5$
- World price
  - $D (2) = 16 - 2 = 14$
  - $S (2) = 2$
- Find efficiency loss
  - Since symmetry
  - $b h = (5 - 2) (5 - 2) = 9$ is our efficiency loss
  - Size of efficiency loss will be greatest when both supply and demand are both very elastic
- Points
  - $(8, 8)$
  - $(11, 5)$
  - $(5, 5)$
  - $(14, 2)$
  - $(2, 2)$

    left=-2; right=17;
    top=17; bottom=-2;
    ---
    x=16-y
    y=x
    (8,8)
    y=5|dashed
    y=2|dashed
    (11,5)
    (5,5)
    (14,2)
    (2,2)

Week 5

Tutorial

ECO101 - Week 05 Failure.pdf

1
- $A$ and $B$ want to purchase a public good
- Their marginal benefits
  - $M B_{A} = 4 - 0.5 q$
  - $M B_{B} = 2 - 0.5 q$
  - $p = 3$
- How much should each roommate spend?
  - Inverse each function
    - $p = 4 - 0.5 q$
      - $q = - 2 p + 8$
      - $0 = - 2 p + 8$
      - $p = 4$
    - $p = 2 - 0.5 q$
      - $q = 4 - 2 p$
      - $0 = 4 - 2 p$
      - $p = 2$
    - Points
      - $(0, 4)$
      - $(0, 2)$
  - Total Benefit
    - $M B = M B_{A} + M B_{B}$
    - $M B = (4 - 0.5 q) + (2 - 0.5 q) = 6 - q$
    - $p = 6 - q$
      - $p - 6 = - q$
      - $- p + 6 = q$
      - $p = 6$
      - $(0, 6)$
    - $M B = M B_{A}$
    - $6 - q = 4 - 0.5 q$
    - $q = 4$
  - $M C = M B$
    - $3 = 6 - q$
    - $q = 3 ⟹ p = 3$
  - $M C = p$
  - $p = 3$
  - Optimal quantity is $3$
  - $3 \cdot 3 = 9$
  - Split on $9$
    - $M B_{A} (3) = 4 - 0.5 (3) = 2.5$
    - $M B_{B} (3) = 2 - 0.5 (3) = 0.5$
    - $\frac{2.5}{3} (9) = 7.5$ roommate $A$
    - $\frac{0.5}{3} (9) = 1.5$ roommate $B$
```
	left=-1; right=9;
	top=7; bottom=-1;
	---
	y=4-0.5x
	y=2-0.5x
	y=6-x
```
2
- Negative externality
- $M B = 100 - 0.25 q$
- Private costs
- $M C = 40 + 0.25 q$
- $M S C = 40 + 0.75 q$
  - $M B = M C$
    - $100 - 0.25 q = 40 + 0.25 q$
    - $q = 120.0 ⟹ p = 70$
  - $M B = M S C$
    - $100 - 0.25 q = 40 + 0.75 q$
    - $q = 60.0 ⟹ p = 85$
- How can we shift $M C ↑$
  - $M C (60) + t = 85$
  - $40 + 0.25 (60) + t = 85$
  - $t = 30.0$
  - $40 + 30 + 0.25 q = 0.25 q + 70$
  - Or
  - $t = M S C (60) - M C (60)$
  - $t = 40 + 0.75 (60) - 40 - 0.25 (60)$
  - $t = 30.0$
- Efficiency Loss
  - $\frac{b h}{2} = \frac{(120 - 60) (100 - 70)}{2} = 900$
```
    left=-5; right=405;
    top=105; bottom=-5;
    ---
    f(x)=100-0.25x
    g(x)=40+0.25x
    h(x)=40+0.75x
    j(x)=0.25x+70
    (120,70)
    (60,85)
    x=120|dashed|0<=y
    x=60|dashed|0<=y
```
3
- $n = 40$
- $M C_{G O} = 20$
- $M C_{Q E W} = 10 + 0.5 D$
- People going on the go is $40 - D$
- $t (40 - D) = 20$
- $t (d) = 10 + 0.5 d$
- $M C_{G O} = M C_{Q E W}$
- $20 = 10 + 0.5 D$
- $D = 20.0$
- This means that we have $20$ on the go and $20$ on the QEW.
- $20 (20) + 20 (10 + 0.5 (20)) = 800.0$ minutes
- Cooperative
  - $M C = 20 + 10 + 0.5 D = 0.5 D + 30$
  - $t = (40 - d) (20) + d (10 + 0.5 d) = - 20 d + d (0.5 d + 10) + 800 = 0.5 d^{2} - 10.0 d + 800.0$
  - $0.5 d^{2} - 10 d + 800$
  - $\frac{\differential t}{\differential d} (0.5 d^{2} - 10 d + 800)$
  - $d - 10 ⟹ d = 10$
  - $t = 0.5 (10)^{2} - 10 (10) + 800 = 750.0$
4
- Positive externality
- $M B = 80 - 0.75 Q$
- $S M B = 80 - 0.25 Q$
- $M C = 20 + 0.25 Q$
- Initial
  - $M B = M C$
  - $80 - 0.75 Q = 20 + 0.25 Q$
  - $Q = 60.0 ⟹ p = 35$
- Social
  - $S M B = M C$
  - $80 - 0.25 Q = 20 + 0.25 Q$
  - $Q = 120.0 ⟹ p = 50$
- How to shift $M B ↑$ towards $S M B$
- $M B (120) + s = 50$
- $(80 - 0.75 (120)) + s = 50$
- $s = 60.0$
- $M B_{s} = 80 + 60 - 0.75 Q = 140 - 0.75 Q$
```
    left=-2; right=200;
    top=200; bottom=-2;
    ---
    y=140-0.75x
    y=80-0.75x
    y=80-0.25x
    y=20+0.25x
```

Week 6

Lecture

ECO101 Lecture 06

$E = 100 - 2 A$
$C = 20 A + A^{2}$
- $M C = 20 + 2 A$
$B = 100 A - A^{2}$
- $M B = 100 - 2 A$

M B = M C

$100 - 2 A = 20 + 2 A$
$A = 20 ⟹ E = 60$

    left=-2; right=105;
    top=105; bottom=-2;
    ---
    y=100-2x
    y=20+2x
    y=20x+x^2
    y=100x-x^2
    (20,60)

Tutorial

ECO101- Week 06 Environmental.pdf

1
- Perfectly competitive market
- Receive a price of $20$ for output
- $M C = 4 Q$
- This creates pollution so $M S C = 5 Q$
- $M B = 20$
- Private
  - $M C = M B$
  - $4 Q = 20$
  - $Q = 5$
- Social
  - $M S C = M B$
  - $5 Q = 20$
  - $Q = 4$
- We want to shift $M C$ towards $M S C$
  - $M C (4) + t = 20$
  - $((4) (4)) + t = 20$
  - $t = 4$
  - $M C_{t} = 4 Q + 4$
```
    left=-2; right=10;
    top=30; bottom=-2;
    ---
    y=20
    y=4x
    y=4x+4|dashed
    y=5x
```
2
- $M B_{1} = 200$
- $C_{1} = 4 q_{1}^{2} + q_{2}^{2} + 20 q_{2}$
- $M B_{2} = 80$
- $C_{2} = 2 q_{2}^{2}$
- 1
  - $M C_{1} = 8 q_{1}$
  - $M B_{1} = M C_{1}$
  - $8 q_{1} = 200$
  - $q_{1} = 25$
- 2
  - $M C_{2} = 4 q_{2}$
  - $M C_{2} = M B_{2}$
  - $80 = 4 q_{2}$
  - $q_{2} = 20$
3
- Two polluting firms
- $C_{1} = 2 A_{1}^{2}$
  - $M C A_{1} = 4 A_{1}$
- $C_{2} = A_{2}^{2}$
  - $M C A_{2} = 2 A_{2}$
- $A_{1} = A_{2} = 30$ units of abatement
- Total we need $60$ units of abatement.
- $M C A_{1} = M C A_{2}$
- $4 A_{1} = 2 A_{2}$
- $A_{2} = 2 A_{1}$
- $A_{1} + A_{2} = 60$
- $A_{1} + 2 A_{1} = 60$
- $A_{1} = 20 ⟹ A_{2} = 40$
- $M C A_{1} (20) = (4) (20) = 80$
- $M C A_{2} (40) = (2) (40) = 80$
  - $80 + 80 = 160$
- Price is $80$
- $C_{1} = (2) (20)^{2} = 800$
- $C_{2} = (40)^{2} = 1600$
- $800 + 1600 = 2400$
4
- $F$ and $G$
- $C_{F} = 2 A_{F}^{2} + 20$
- $C_{G} = A_{G}^{2} + 8$
- $B = 14 A - 0.5 A^{2} + 20$
  - $M B = 14 - A$
- $M C_{F} = 4 A_{F}$
  - $\frac{1}{4} M C_{F} = A_{F}$
- $M C_{G} = 2 A_{G}$
  - $\frac{1}{2} M C_{G} = A_{G}$
- Calculate Optimal Abatement
  - $A = A_{F} + A_{G}$
  - $A = \frac{1}{4} M C_{F} + \frac{1}{2} M C_{G}$
  - $A = \frac{1}{4} M C + \frac{1}{2} M C$
  - $A = \frac{3}{4} M C$
  - $\frac{4}{3} A = M C$
  - $M C = M B$
    - $14 - A = \frac{4}{3} A$
    - $A = 6$
    - $A = A_{F} + A_{G}$
    - $A_{F} = 3$
    - $A_{G} = 3$
- Calculate Split
  - $A_{G} = 2 A_{F}$
  - $6 = A_{G} + A_{F}$
  - $6 = 2 A_{F} + A_{F}$
  - $A_{F} = 2 ⟹ A_{G} = 4$
  - Cost before and after
  - $C_{F} (2) = (2) (2)^{2} + 20 = 28$
  - $C_{G} (4) = (4)^{2} + 8 = 24$
    - $28 + 24 = 52$
  - $C_{F} (3) = (2) (3)^{2} + 20 = 38$
  - $C_{G} (3) = (3)^{2} + 8 = 17$
    - $38 + 17 = 55$
  - It's better overall.

Week 7

ECO101 - Week 07 Household Demand.pdf

Tutorial

1
- $u (x, y) = \sqrt{x} \sqrt{y}$
- $p_{x} = 0.5$
- $p_{y} = 0.25$
- $m_{0} = 40$
- $m_{1} = 80$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{X}$
    - $\frac{y}{x} = \frac{p_{x}}{p_{y}}$
    - $0.25 y = 0.5 x$
    - $y = 2.0 x$
    - $x = 0.5 y$
  - Budget Line
    - $p_{x} x + p_{y} y = m_{0}$
    - $0.5 x + 0.25 y = 40$
    - $0.5 x + 0.25 (2 x) = 40$
    - $x = 40.0 ⟹ y = 20.0$
  - Utility
    - $u (x, y) = \sqrt{x} \sqrt{y}$
    - $u (x, y) = \sqrt{40} \sqrt{20}$
- Bundle C
  - Tangency
    - $y = 2.0 x$
    - $x = 0.5 y$
  - Budget Line
    - $p_{x} x + p_{y} y = m_{1}$
    - $0.5 x + 0.25 y = 80$
    - $0.5 x + 0.25 (2 x) = 80$
    - $x = 80.0 ⟹ y = 40.0$
  - Utility
    - $u (x, y) = \sqrt{80} \sqrt{40}$
- Bundle B only if change in price, none, so no bundle B.
```
    left=-2; right=100;
    top=100; bottom=-2;
    ---
    xy=3200
    xy=800
    0.5x+0.25y=80
    0.5x+0.25y=40
```
2
- $u (x, y) = \sqrt{x} \sqrt{y}$
- $m = 60$
- $p_{y_{0}} = 0.5$
- $p_{y_{1}} = 0.5$
- $p_{x_{0}} = 1$
- $p_{x_{1}} = 0.25$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{X}$
    - $\frac{y}{x} = \frac{p_{x_{0}}}{p_{y_{0}}}$
    - $0.5 y = x$
    - $y = 2 x$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $x + 0.5 y = 60$
    - $x + 0.5 (2 x) = 60$
    - $x = 30.0 ⟹ y = 60.0$
  - Utility
    - $u (x, y) = \sqrt{x} \sqrt{y}$
    - $u (x, y) = \sqrt{x y}$
    - $u (x, y) = \sqrt{1800}$
- Bundle C
  - Tangency
    - $M R S = O C O S T_{X}$
    - $\frac{y}{x} = \frac{p_{x_{1}}}{p_{y_{1}}}$
    - $0.5 y = 0.25 x$
    - $x = 2.0 y$
    - $y = 0.5 x$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $0.25 x + 0.5 y = 60$
    - $0.25 x + 0.5 (0.5 x) = 60$
    - $x = 120.0 ⟹ y = 60.0$
  - Utility
    - $u (x, y) = \sqrt{x y}$
    - $u (x, y) = \sqrt{7200}$
- Bundle B
  - Utility of A and price of C
  - $x y = 1800$
  - $y = 0.5 x$
  - $0.5 x^{2} = 1800$
  - $x = - 60.0 \lor x = 60.0$
  - $x = 60 ⟹ y = 30$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $\frac{1}{4} x + \frac{1}{2} y = m$
    - $\frac{1}{4} (60) + \frac{1}{2} (30) = m$
    - $m = 30$
    - $\frac{1}{4} x + \frac{1}{2} y = 30$
3
- $u (x, y) = \sqrt{x} \sqrt{y}$
- $M R S = \frac{y}{x}$
- $m = 600$
- $p_{y} = 2$
- We want $x = 10$
- TBU
- Tangency
  - $M R S = O C O S T_{X}$
  - $\frac{y}{x} = \frac{p_{x}}{p_{y}}$
  - $2 y = 10 p_{x}$
  - $y = 5 p_{x}$
- Budget Line
  - $p_{x} x + p_{y} y = m$
  - $10 p_{x} + (2) (5 p_{x}) = 600$
  - $p_{x} = 30$
- Utility
  - $u (x, y) = \sqrt{x y}$
  - $u (x, y) = \sqrt{(10) ((5) (30))}$
  - $u (x, y) = \sqrt{1500}$
```
    left=-2; right=100;
    top=500; bottom=-2;
    ---
    30x+2y=600
    xy=1500
    (10,5*30)
```
4
- $u (x, y) = \sqrt{x} \sqrt{y}$
- $m = 80$
- $p_{x_{0}} = 1$
- $p_{y_{0}} = 0.5$
- $p_{x_{1}} = 0.25$
- $p_{y_{1}} = 0.5$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{X}$
    - $\frac{y}{x} = \frac{p_{x_{0}}}{p_{y_{0}}}$
    - $0.5 y = x$
    - $y = 2 x$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $x + 0.5 y = 80$
    - $x + 0.5 (2 x) = 80$
    - $x = 40.0 ⟹ y = 80.0$
  - Utility
    - $x y = 3200$
- Bundle C
  - Tangency
    - $M R S = O C O S T_{X}$
    - $0.5 y = 0.25 x$
    - $y = 0.5 x$
  - Budget Line
    - $0.25 x + 0.5 y = 60$
    - $0.25 x + 0.5 (0.5 x) = 60$
    - $x = 120.0 ⟹ y = 60.0$
  - Utility
    - $x y = 7200$
- Bundle B
  - Utility A, tangency C
  - $x y = 3200$
  - $y = 0.5 x$
  - $0.5 x^{2} = 3200$
  - $x = - 80.0 \lor x = 80.0$
  - $x = 80 ⟹ y = 40$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $0.25 x + 0.5 y = m$
    - $0.25 (80) + 0.5 (40) = m$
    - $m = 40.0$
- Cost to join this price club?

Week 8

Tutorial

ECO101 - Week 08 Household Supply.pdf

1
- $u (f_{1}, f_{2}) = \sqrt{f_{1} f_{2}}$
- $M R S = \frac{f_{2}}{f_{1}}$
- $m_{0} = 80$
- $m_{1} = 50$
- $r = 0.25$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{Z}$
    - $\frac{f_{2}}{f_{1}} = 1 + r$
    - $\frac{f_{2}}{f_{1}} = 1.25$
    - $1.25 f_{1} = f_{2}$
    - $f_{1} = 0.8 f_{2}$
  - Budget Line
    - $f_{1} + \frac{1}{1 + r} f_{2} = m_{0} + \frac{1}{1 + r} m_{1}$
    - $f_{1} + \frac{f_{2}}{1.25} = 80 + \frac{50}{1.25}$
    - $f_{1} + \frac{1.25 f_{1}}{1.25} = 80 + \frac{50}{1.25}$
    - $f_{1} = 60 ⟹ f_{2} = 75$
  - Utility
    - $u (f_{1}, f_{2}) = \sqrt{f_{1} f_{2}}$
    - $u (f_{1}, f_{2}) = \sqrt{4500}$
- Savings
  - $s = m_{1} - f_{1}$
  - $s = 80 - 60 = 20$
```
    left=-2; right=200;
    top=200; bottom=-2;
    ---
    xy=4500
    x+0.8y=120
    (60,75)
    (80,50)
```
2
- $u (z, f) = \sqrt{z f}$
- $24 - z = h$
- $w = 5$
- $A = 60$
- Tangency
  - $M R S = O C O S T_{Z}$
  - $\frac{f}{z} = w$
  - $f = 5 z$
- Budget Line
  - $f (z) = A + w (24 - z)$
  - $f (z) = 60 + (5) (24 - z) = 180 - 5 z$
  - $5 z = 180 - 5 z$
  - $z = 18 ⟹ f = 90$
- Utility
  - $u (z, f) = \sqrt{1620}$
```
    left=-2; right=50;
    top=200; bottom=-2;
    ---
    f(z)=180-5z
    xy=1620
    (18,90)
```
3
- $u (c_{1}, c_{2}) = \sqrt{c_{1} c_{2}}$
- $m_{1} = 60$
- $m_{2} = 40$
- $r = 0.25$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{C_{2}}$
    - $\frac{c_{2}}{c_{1}} = 1 + r$
    - $\frac{c_{2}}{c_{1}} = 1.25$
    - $c_{2} = 1.25 c_{1}$
  - Budget Line
    - $c_{1} + \frac{1}{1.25} c_{2} = m_{1} + \frac{1}{1.25} m_{2}$
    - $c_{1} + \frac{1}{1.25} c_{2} = (60) + \frac{1}{1.25} (40) = 92.0$
    - $c_{1} + \frac{1}{1.25} c_{2} = 92$
    - $c_{1} + \frac{1}{1.25} 1.25 c_{1} = 92$
    - $c_{1} = 46.0 ⟹ c_{2} = 57.5$
  - Utility
    - $u (c_{1}, c_{2}) = \sqrt{2645}$
- $s = m_{1} - c_{1}$
  - $s = 60 - 46 = 14$ are saved
```
    left=-2; right=100;
    top=100; bottom=-2;
    ---
    xy=2645
    x+0.8y=92
    (46,57.5)
    (60,40)
```
4
- $u (z, f) = \sqrt{z f}$
- $24 = z + h$
- $A = 6$
- $w = 0.5$
- Tangency
  - $M R S = O C O S T_{Z}$
  - $\frac{f}{z} = w$
  - $f = 0.5 z$
- Budget Line
  - $f (z) = A + w (24 - z)$
  - $f (z) = 6 + 0.5 (24 - z) = 18 - 0.5 z$
  - $0.5 z = 18 - 0.5 z$
  - $z = 18.0 ⟹ f = 9$
- Utility
  - $u (z, f) = \sqrt{162}$
- $z = 18 ⟹ h = 6$
```
    left=-2; right=40;
    top=20; bottom=-2;
    ---
    xy=162
    f(z)=18-0.5z
    (18,9)
```

Week 9

Tutorial

ECO101 - Week 09 Short Run Firms.pdf

1
- Profit maximizing firm
- $S V C = q^{3} - 40 q^{2} + 1400 q$
- $S F C = 4000$
- $S M C = 3 q^{2} - 80 q + 1400$
- Calculate shutdown price
- This is when your $S M C < S A V C$
- $S A V C = q^{2} - 40 q + 1400$
- When is $S M C = S A V C$
  - $3 q^{2} - 80 q + 1400 = q^{2} - 40 q + 1400$
  - $q = 0 \lor q = 20$
  - $p = S M C$
  - $p = 3 q^{2} - 80 q + 1400$
  - $p = (3) (20)^{2} - (80) (20) + 1400$
  - $p = 1000$
- $S A V C = q^{2} - 40 q + 1400$
- $S A C = S A V C + S A F C$
- $S A C = (q^{2} - 40 q + 1400) + (4000 q^{- 1}) = q^{2} - 40 q + 1400 + 4000 \frac{1}{q}$
- $S M C = 3 q^{2} - 80 q + 1400$
```
    left=-5; right=100;
    top=2000; bottom=-5;
    ---
    y=x^2-40x+1400
    y=x^2-40x+1400+4000x^{-1}
    y=3x^2-80x+1400
    (20,1000)
```
2
- $q = 2 \sqrt{k} \sqrt{l}$
- $w_{0} = 2$
- $w_{1} = 1$
  $k = 16$
- $r = 1$
- $p = 4$
- Find new labour
- Bundle A
  - $q = 2 \sqrt{16} \sqrt{l} = 8 \sqrt{l}$
  - $q = 8 \sqrt{l}$
  - $l = \frac{1}{64} q^{2}$
  - $S T C = w l + k r$
  - $S T C = \frac{1}{32} q^{2} + 16$
  - $S M C = \frac{1}{16} q$
  - Max profits at $p = S M C$
  - $p = \frac{1}{16} q$
  - $4 = \frac{1}{16} q$
  - $q = 64$
  - $l = \frac{1}{64} 64^{2} = 64$
- Bundle B
  - $q = 8 \sqrt{l}$
  - $l = \frac{1}{64} q^{2}$
  - $S T C = w l + k r$
  - $S T C = l + 16$
  - $S T C = \frac{1}{64} q^{2} + 16$
  - $S M C = \frac{1}{32} q$
  - $p = S M C$
  - $\frac{1}{32} q = p$
  - $\frac{1}{32} q = 4$
  - $q = 128$
  - $l = \frac{1}{64} (128)^{2} = 256$
- $l = 64 \to 256 ⟹ l ↑ (192)$
- Profit
  - $π_{a} = R - C$
  - $π_{a} = R - S T C$
  - $π_{a} = P Q - S T C$
  - $π_{a} = 4 q - \frac{1}{32} q^{2} - 16 = - \frac{q^{2}}{32} + 4 q - 16$
  - $π_{b} = R - C$
  - $π_{b} = 4 q - \frac{1}{64} q^{2} - 16 = - \frac{q^{2}}{64} + 4 q - 16$
```
    left=-2; right=300;
    top=200; bottom=-2;
    ---
	y=1/2 x+ 32
	y=1/4 x+ 64
	y=8\sqrt{x}
```
4
- Price taking firm
- $q = \sqrt{k} \sqrt{l}$
- $k = 16$
- $q = 4 \sqrt{l}$
- Isolate $l$
- $l = \frac{q^{2}}{16}$
- $w = 4$
- $r = 4$
- $p_{0} = 2$
- $p_{1} = 8$
- Bundle A
  - $S T C = w l + r k$
  - $S T C = \frac{1}{4} q^{2} + 64$
  - $S M C = \frac{1}{2} q$
  - $p = S M C$
  - $2 = \frac{1}{2} q$
  - $q = 4$
  - $4 = 4 \sqrt{l}$
  - $l = 1$
- Bundle B
  - $S T C = \frac{1}{4} q^{2} + 64$
  - $S M C = \frac{1}{2} q$
  - $p = S M C$
  - $8 = \frac{1}{2} q$
  - $q = 16$
  - $16 = 4 \sqrt{l}$
  - $l = 16$
- Isoprofit lines
  - $π_{a} : R - C$
  - $π_{a} : p q - (w l + r k)$
    - $(2) (4) - ((4) (1) + (4) (16)) = - 60$
    - $p q = π + w l + r k$
    - $q = \frac{π + w l + r k}{p}$
    - $q = \frac{w}{p} l + \frac{π + r k}{p}$
    - $q = \frac{4}{2} l + \frac{(- 60) + (4) (16)}{2}$
    - $q = 2 l + 2$
  - $π_{b} = p q - (w l + r k)$
    - $π_{b} = (8) (16) - ((4) (16) + (4) (16))$
    - $π_{b} = 0$
    - $q = \frac{w}{p} l + \frac{π + r k}{p}$
    - $q = \frac{4}{8} l + \frac{(4) (16)}{8}$
    - $q = \frac{l}{2} + 8$
```
    left=-2; right=30;
    top=30; bottom=-2;
    ---
	y=1/2 x+8
	y=2x+2
	y=4\sqrt{x}
	(1,4)
	(16,16)
```

Week 10

Lecture

ECO101 Lecture 10

Tutorial

ECO101 - Week 10 Long Run Firms.pdf

1
- $q = \sqrt[4]{k} \sqrt[4]{l}$
- $w_{0} = \frac{1}{8}$
- $w_{1} = \frac{1}{2}$
- $r = \frac{1}{8}$
- $p = 3$
- Bundle A
  - Tangency
    - $M R T S = O C O S T$
    - $\frac{k}{l} = \frac{w}{r}$
    - $\frac{1}{8} k = \frac{1}{8} l$
    - $k = l$
  - Sufficiency
    - $q = \sqrt[4]{k} \sqrt[4]{l}$
    - $q = \sqrt[4]{l} \sqrt[4]{l}$
    - $q = \sqrt[4]{l^{2}}$
    - $q = l^{\frac{2}{4}}$
    - $q = l^{\frac{1}{2}}$
    - $q = \sqrt{l}$
    - $q^{2} = l$
  - $S T C = w l + r k$
  - $S T C = \frac{1}{8} q^{2} + \frac{1}{8} q^{2} = \frac{q^{2}}{4}$
  - $S M C = \frac{1}{2} q$
  - Optimize
  - $p = \frac{1}{2} q$
  - $3 = \frac{1}{2} q$
  - $q = 6$
  - $6^{2} = l$
  - $l = 36$
  - Find cost
    - $S T C = \frac{6^{2}}{4} = 9$
  - $9 = w l + r k$
  - $9 = \frac{1}{8} l + \frac{1}{8} k$
  - $9 = \frac{1}{8} l + \frac{1}{8} k$
  - $k = 72 - l$
  - Utility
    - $u (k, l) = \sqrt[4]{1296}$
- Bundle B
  - Tangency
    - $M R T S = O C O S T$
    - $\frac{k}{l} = \frac{w}{r}$
    - $\frac{1}{8} k = \frac{1}{2} l$
    - $k = 4 l$
    - $\frac{1}{4} k = l$
  - Sufficiency
    - $q = \sqrt[4]{k} \sqrt[4]{l}$
    - $q = \sqrt[4]{4 l} \sqrt[4]{l}$
    - $q = \sqrt[4]{4 l^{2}}$
    - $q = 4 l^{\frac{2}{4}}$
    - $q^{2} = 4 l$
    - $\frac{1}{4} q^{2} = l$
  - $S T C = w l + r k$
  - $S T C = \frac{1}{2} \frac{1}{4} q^{2} + \frac{1}{8} q^{2}$
  - $S T C = \frac{q^{2}}{4}$
  - $S M C = \frac{1}{2} q$
  - $3 = \frac{1}{2} q$
  - $q = 6$
  - $6 = \sqrt{4 l}$
  - $l = 9$
  - $\frac{1}{4} k = 9$
  - $k = 36$
  - $S T C = \frac{q^{2}}{4}$
  - $S T C = \frac{6^{2}}{4}$
  - $S T C = 9$
  - $9 = w l + r k$
  - $9 = \frac{1}{2} l + \frac{1}{8} k$
  - $9 = \frac{1}{2} l + \frac{1}{8} k$
  - $k = 72 - 4 l$
  - Utility
    - $u (k, l) = \sqrt[4]{324}$
```
    left=-2; right=80;
    top=80; bottom=-2;
    ---
    y=72-4x
    y=72-x
    xy=324
    xy=1296
    (36,36)
    (9,36)
```
2
- Cost minimizing firm
- $q = \sqrt{k} \sqrt{l}$
- $w = 2$
- $r = 2$
- $q_{0} = 2$
- $q_{1} = 4$
- Short run costs
  - $q^{2} = k l$
  - $\frac{q^{2}}{l} = k$
  - $\frac{q^{2}}{k} = l$
  - Bundle A
    - Tangency Condition
      - $M R T S = O C O S T_{L}$
      - $\frac{k}{l} = \frac{w}{r}$
      - $2 l = 2 k$
      - $k = l$
    - Sufficiency
      - $2^{2} = l^{2}$
      - $2 = l = k$
    - Budget Line
      - $C = w l + r k$
      - $C = (2) (2) + (2) (2) = 8$
      - $8 = 2 l + 2 k$
      - $k = 4 - l$
    - Utility
      - $u (k, l) = \sqrt{k l}$
      - $u (k, l) = \sqrt{4} = 2$
  - Bundle B
    - Tangency
      - $M R T S = O C O S T$
      - $w l = r k$
      - $2 l = 2 k$
      - $l = k$
    - Sufficiency
      - $q = \sqrt{k} \sqrt{l}$
      - $q^{2} = k l$
      - $4^{2} = k l$
      - Since $k$ was $2$
      - $4^{2} = 2 l$
      - $l = 8$
    - Budget Line
      - $C = w l + r k$
      - $C = (2) (8) + (2) (2) = 20$
      - $20 = 2 l + 2 k$
      - $k = 10 - l$
    - Utility
      - $u (k, l) = \sqrt{k l}$
      - $u (k, l) = \sqrt{16} = 4$
```
	left=-2; right=20;
	top=20; bottom=-2;
	---
	xy=4
	xy=2
	y=10-x
	y=8-x
	y=4-x
```
3
- Cost minimizing business
- $q = \sqrt{k} \sqrt{l}$
- $w_{0} = 4$
- $w_{1} = 16$
- $r = 4$
- $q = 300$
- Short run, since capital is fixed
- Bundle A
  - Tangency
    - $M R T S = O C O S T$
    - $w l = r k$
    - $4 l = 4 k$
    - $k = l$
  - Sufficiency
    - $q^{2} = k l$
    - $q^{2} = l^{2}$
    - $q = l = k$
    - $300 = l = k$
  - Costs
    - $C = w l + r k$
    - $C = 4 l + 4 k$
    - $C = (4) (300) + (4) (300) = 2400$
    - $2400 = 4 l + 4 k$
    - $k = 600 - l$
  - Utility
    - $u (k, l) = \sqrt{90000}$
- Bundle B
  - Same Capital
  - Tangency
    - $w l = r k$
    - $16 l = 4 k$
    - $k = 4 l$
    - $\frac{1}{4} k = l$
  - Sufficiency
    - $q^{2} = k l$
    - $q^{2} = 4 l^{2}$
    - $q = 2 l$
    - $q = (2) (\frac{1}{4} k) = \frac{1}{2} k$
    - $300 = \frac{1}{2} k$
    - $k = 600$
    - $300 = 2 l$
    - $l = 150$
  - Costs
    - $C = w l + r k$
    - $C = (16) (150) + (4) (600) = 4800$
    - $4800 = 16 l + 4 k$
    - $k = 1200 - 4 l$
  - Utility
    - $u (k, l) = \sqrt{90000}$
```
    left=-2; right=1500;
    top=1500; bottom=-2;
    ---
    (300,300)
    (150,600)
    y=1200-4x
    y=600-x
    xy=90000
```
4
- $q = \sqrt[4]{k} \sqrt[4]{l}$
- $w_{0} = \frac{1}{8}$
- $w_{1} = \frac{1}{2}$
- $r = \frac{1}{8}$
- $p = 3$
- Long run
- Bundle A
  - Tangency
    - $M R T S = O C O S T$
    - $\frac{k}{l} = \frac{w}{r}$
    - $w l = r k$
    - $\frac{1}{8} l = \frac{1}{8} k$
    - $l = k$
  - Sufficiency
    - $q^{4} = k l$
    - $q^{4} = l^{2}$
    - $q^{2} = l = k$
  - Cost
    - $C = w l + r k$
    - $C = \frac{1}{8} l + \frac{1}{8} k$
    - $C = \frac{1}{8} l + \frac{1}{8} k$
    - $C = \frac{1}{8} l + \frac{1}{8} l = \frac{l}{4} = \frac{q^{2}}{4}$
    - $M C = \frac{1}{2} q$
  - $p = M C$
    - $p = \frac{1}{2} q$
    - $3 = \frac{1}{2} q$
    - $q = 6$
    - $6^{2} = l$
    - $l = 36 = k$
  - Budget Line
    - $C = \frac{1}{8} l + \frac{1}{8} k$
    - $C = \frac{1}{8} (36) + \frac{1}{8} (36) = 9$
    - $9 = \frac{1}{8} l + \frac{1}{8} k$
    - $k = 72 - l$
  - Utility
    - $u (k, l) = \sqrt{1296}$
- Bundle B
  - Tangency
    - $w l = r k$
    - $\frac{1}{2} l = \frac{1}{8} k$
    - $k = 4 l$
    - $l = \frac{1}{4} k$
  - Sufficiency
    - $q^{4} = k l$
    - $q^{4} = 4 l^{2}$
    - $q^{2} = 2 l$
    - $\frac{1}{2} q^{2} = l$
    - $\frac{1}{2} q^{2} = \frac{1}{4} k$
    - $k = 2 q^{2}$
  - Cost
    - $C = w l + r k$
    - $C = \frac{1}{2} l + \frac{1}{8} k$
    - $C = \frac{1}{2} \frac{1}{2} q^{2} + \frac{1}{8} 2 q^{2}$
    - $C = \frac{q^{2}}{2}$
    - $M C = q$
  - $p = M C$
    - $3 = q$
    - $k = (2) (3)^{2} = 18$
    - $l = \frac{1}{2} (3)^{2} = \frac{9}{2}$
  - Budget Line
    - $C = \frac{1}{2} l + \frac{1}{8} k$
    - $C = \frac{1}{2} (\frac{9}{2}) + \frac{1}{8} (18) = \frac{9}{2}$
    - $\frac{9}{2} = \frac{1}{2} l + \frac{1}{8} k$
    - $k = 36 - 4 l$
  - Utility
    - $u (k, l) = \sqrt{k l}$
    - $u (k, l) = \sqrt{81}$
- So initially she will hire $36$ workers, now she'll hire $\frac{9}{2}$ .
```
    left=-2; right=80;
    top=80; bottom=-2;
    ---
    xy=81
    xy=1296
    y=36-4x
    y=72-x
    (9/2,18)
    (36,36)
```

Week 11

Lecture

1
- $q = \sqrt[4]{k} \sqrt[4]{l}$
- $w = 0.5$
- $r = 0.5$
- $F = 4$
- $Q = 160 - 20 p$
- a
  - Find long run cost
  - Tangency
    - $k = l$
  - Sufficiency
    - $q^{4} = k l$
    - $l = q^{2} = k$
  - Cost
    - $C = \frac{1}{2} l + \frac{1}{2} k$
    - $C = \frac{1}{2} q^{2} + \frac{1}{2} q^{2} = q^{2}$
    - $C = q^{2} + F$
    - $C = q^{2} + 4$
    - $M C = 2 q$
    - $A C = q + 4 q^{- 1}$
  - MES
    - $M C = A C$
    - $q + 4 q^{- 1} = 2 q$
    - $q = - 2 \lor q = 2$
    - $q_{0} = 2$
  - $p = M C$
    - $p = 2 q_{0}$
    - $p_{0} = 2 (2) = 4$
  - Find quantity of industry
    - $Q_{0} = 160 - (20) (p_{0})$
    - $Q_{0} = 160 - (20) (4) = 80$
  - Find firms
    - $\frac{Q_{0}}{q_{0}} = n_{0}$
    - $\frac{80}{2} = 40$
- b
  - Industry Supply
    - $p = M C$
    - $p = 2 q$
    - $\frac{1}{2} p = q$
    - So
    - $q_{s} = \frac{1}{2} p$
    - $Q_{s} = n \cdot q_{s}$
    - $Q_{s} = 40 \cdot \frac{1}{2} p$
    - $Q_{s} = 20 p$
- c
  - What if demand is now $D_{1} = 320 - 20 p$
  - Bundle A
    - We have the same amount of firms
    - $n_{0} = 40$
    - Tangency
      - $M R T S = O C O S T$
      - $w l = r k$
      - $\frac{1}{2} l = \frac{1}{2} k$
      - $k = l$
    - Sufficiency
      - $q^{4} = k l$
      - $q^{2} = l = k$
    - Cost
      - $C = \frac{1}{2} l + \frac{1}{2} k$
      - $C = \frac{1}{2} q^{2} + \frac{1}{2} q^{2} = q^{2} + 4$
      - $M C = 2 q$
      - $A C = q + 4 q^{- 1}$
    - MES
      - $M C = A C$
      - $2 q = q + 4 q^{- 1}$
      - $q = - 2 \lor q = 2$
      - $q_{1} = 2$
    - $p = M C$
      - $p_{0} = 2 (2) = 4$
    - Quantity of industry
      - $Q_{1} = 320 - (20) (4) = 240$
    - Quantity of firms
      - $n_{1} = \frac{Q_{1}}{q_{1}}$
      - $n_{1} = \frac{240}{2} = 120$
    - Industry Supply
      - $p = M C$
      - $p = 2 q$
      - $\frac{1}{2} p = q$
      - $Q_{s} = n_{1} \frac{1}{2} p$
      - $Q_{s} = 60 p$
1
- $q = \sqrt[4]{k} \sqrt[4]{l}$
- $w = 0.5$
- $r = 0.5$
- $F = 4$
- $Q = 160 - 20 p$
- Bundle A: short run
  - $n_{0} = 40$
  - $Q_{d} = 320 - 20 p$
  - $Q_{s} = 20 p$
  - $20 p = 320 - 20 p$
  - $p = 8$
  - Industry Quantity:
    - $Q = 20 (8) = 160$
  - Firm Quantity:
    - $Q_{s} = n \cdot q$
    - $160 = 40 q$
    - $q = 4$
  - $p = 8$
  - $Q = 160$
  - $q = 4$
  - $π = T R - T C$
    - $π = (8 \cdot 4) - (4^{2} + 4) = 12$
- Bundle B: Long run
  - $p = (M C = A C)$
  - $M C = A C ⟹ p = 4$
  - MES $⟹ q = 2$
  - $Q = 320 - 20 (4) = 240$
  - Firms:
    - $n_{1} = \frac{Q_{1}}{q_{1}}$
    - $n_{1} = \frac{240}{2} = 120$
  - $p = 4$
  - $Q = 240$
  - $q = 2$
  - $n_{1} = 120$
2
- $C = q^{2} + 4$
- $Q = 160 - 20 p$
- $D_{1} = 120 - 20 p$
- $n_{0} \to n_{1}$
- Bundle A
  - $M C = 2 q$
  - $A C = q + 4 q^{- 1}$
  - $p = (M C = A C)$
  - $M C = A C$
    - $2 q = q + 4 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 2 q$
    - $p = 2 (2) = 4$
  - Industry Quantity
    - $Q = 160 - 20 (4) = 80$
  - Firm Quantity
    - $Q_{s} = n q$
    - $80 = (n) (2)$
    - $n = 40$
- Bundle B
  - Industry Quantity
    - $Q = 120 - 20 p$
    - $Q = 120 - 20 (4) = 40$
  - Firm Quantity
    - $Q_{s} = n q$
    - $40 = (n) (2)$
    - $n = 20$
3
- $\begin{array}{l} Price & Industry Supply & Firm Supply & MC & AVC & ATC \\ 8 & 400000 & 350 & 7 & 7 & 9 \\ 9 & 350000 & 400 & 8 & 7.10 & 10.43 \\ 10 & 300000 & 450 & 9 & 7.20 & 10.06 \\ 11 & 250000 & 500 & 10 & 7.50 & 10 \\ 12 & 200000 & 550 & 11 & 8 & 10.22 \end{array}$
- In the short run, there are $1000$ firms
  - $n = 1000$
  - Find market price
  - $p = (M C = A C)$
  - $M C = A C$
    - $10 = 10$
  - When do we have $1000$ firms
  - $n = \frac{Q}{q}$
  - $n q = Q$
  - $p = 8$
    - $Q = 400000$
    - $q = 400$
    - $⟹ n = 1000$
  - $p = 10$
    - $Q = 300000$
    - $q = 500$
    - $500 n = 300000$
    - $n = 600$

Tutorial

ECO101 - Week 11 Competition.pdf

1
- $D = 680 - 20 p$
- How many firms will be in the industry?
- $n_{0} = \frac{Q_{0}}{q_{0}}$
- $C = w l + r k$
- $p = M C$
  - $2 = 2$
  - $C = 9$
- $\begin{array}{l} Output & 1 & 2 & 3 & 4 & 5 & 6 \\ Cost & 8 & 9 & 12 & 17 & 24 & 33 \\ MC & 0 & 2 & 4 & 6 & 8 & 10 \\ AC & 8 & 4.5 & 4 & 4.25 & 4.8 & 5.5 \end{array}$
- $p = (M C = A C)$
  - $M C = A C$
    - $q = 3$
    - $4 = 4$
- $p = 4$
- Firms
  - $D = 680 - (20) (4) = 600$
  - $n q = Q$
  - $3 n = 600$
  - $n = 200$
2
- $C = 20 q^{2} + 80$
- $M C = 40 q$
- $A C = 20 q + 80 q^{- 1}$
- $Q = 2400 - 20 p$
- $s = 20$ subsidy
- How many firms will enter?
- Bundle A: $s = 0$
  - $p = (M C = A C)$
  - $M C = A C$
    - $40 q = 20 q + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 (2) = 80$
  - $Q = 2400 - 20 (80) = 800$
  - $n = \frac{Q}{q}$
  - $n = \frac{800}{2} = 400$
  - $p = M C$
    - $p = 40 q$
    - $q = \frac{p}{40}$
  - $Q_{s} = n p$
  - $Q_{s} = (400) (\frac{p}{40}) = 10 p$
- Bundle C: $s = 20$
  - Subsidy reduces costs
  - $C = 20 q^{2} + 80 - 20 q$
  - $M C = 40 q - 20$
  - $A C = 20 q - 20 + 80 q^{- 1}$
  - $p = (M C = A C)$
  - $M C = A C$
    - $40 q - 20 = 20 q - 20 + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 (2) - 20$
    - $p = 60$
  - $Q = 2400 - 20 (60) = 1200$
  - $n = \frac{Q}{q}$
  - $n = \frac{1200}{2} = 600$
  - $p = M C$
    - $p = 40 q - 20$
    - $q = \frac{p}{40} + \frac{1}{2}$
  - $Q_{s} = n p$
    - $Q_{s} = (600) (\frac{p}{40} + \frac{1}{2}) = 15 p + 300$
- Bundle B:
  - $C = 20 q^{2} + 80 - 20 q$
  - $M C = 40 q - 20$
  - $A C = 20 q - 20 + 80 q^{- 1}$
  - $n = 400$
  - $p = M C$
    - $p = 40 q - 20$
    - $q = \frac{p}{40} + \frac{1}{2}$
  - $Q_{s} = n p$
  - $Q_{s} = (400) (\frac{p}{40} + \frac{1}{2}) = 10 p + 200$
  - $Q_{s} = Q_{d}$
  - $10 p + 200 = 2400 - 20 p$
  - $p = \frac{220}{3}$
  - $10 (\frac{220}{3}) + 200 = \frac{2800}{3}$
```
	left=-2; right=2500;
	top=300; bottom=-2;
	---
	x=2400-20y
	x=10y+200
	x=15y+300
	x=10y
	(800,80)
	(1200,60)
	(2800/3,220/3)
```
3
- $C = 20 q^{2} + 80$
- $M C = 40 q$
- $A C = 20 q + 80 q^{- 1}$
- $Q = 3600 - 20 p$
- $t = 20$ tax
- How many firms in both sections
- Bundle A
  - $M C = A C$
    - $40 q = 20 q + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 (2) = 80$
  - $Q = 3600 - 20 (80) = 2000$
  - Firms
    - $n = \frac{Q}{q}$
    - $n = \frac{2000}{2} = 1000$
  - Industry Supply
    - $Q_{s} = n p$
    - $p = M C$
    - $p = 40 q$
    - $\frac{1}{40} p = q$
    - $Q_{s} = (1000) (\frac{1}{40} p) = 25 p$
  - Autarky
    - $Q_{s} = Q_{d}$
    - $25 p = 3600 - 20 p$
    - $p = 80 ⟹ q = 2000$
  - $q$
    - $q = \frac{1}{40} p$
    - $q = \frac{1}{40} (80) = 2$
    - $(2, 80)$
- Bundle C
  - Tax adds to cost
  - $C = 20 q^{2} + 20 q + 80$
  - $A C = 20 q + 20 + 80 q^{- 1}$
  - $M C = 40 q + 20$
  - $M C = A C$
    - $40 q + 20 = 20 q + 20 + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 (2) + 20 = 100$
  - $Q$
    - $Q = 3600 - 20 (100) = 1600$
  - $n$
    - $n = \frac{Q}{q}$
    - $n = \frac{1600}{2} = 800$
  - $Q_{s}$
    - $Q_{s} = n p$
  - $p$
    - $p = M C$
    - $p = 40 q + 20$
    - $q = \frac{p}{40} - \frac{1}{2}$
  - $Q_{s}$
    - $Q_{s} = n p$
    - $Q_{s} = (800) (\frac{p}{40} - \frac{1}{2}) = 20 p - 400$
  - $Q_{s} = Q_{d}$
    - $20 p - 400 = 3600 - 20 p$
    - $p = 100 ⟹ q = 1600$
  - $q$
    - $q = \frac{1}{40} p - \frac{1}{2}$
    - $q = \frac{1}{40} (100) - \frac{1}{2} = 2$
    - $(2, 100)$
- Bundle B
  - $C = 20 q^{2} + 20 q + 80$
  - $A C = 20 q + 20 + 80 q^{- 1}$
  - $M C = 40 q + 20$
  - $n = 1000$
  - $M C = A C ⟹ q = 2$
  - $p = M C ⟹ p = 40 q + 20$
  - $p = M C ⟹ q = \frac{p}{40} - \frac{1}{2}$
  - $Q_{s}$
    - $Q_{s} = n p$
    - $Q_{s} = (1000) (\frac{p}{40} - \frac{1}{2}) = 25 p - 500$
  - $Q_{s} = Q_{d}$
    - $25 p - 500 = 3600 - 20 p$
    - $p = \frac{820}{9}$
    - $Q = 3600 - 20 (\frac{820}{9}) = \frac{16000}{9}$
    - $(\frac{16000}{9}, \frac{820}{9})$
  - $q$
    - $q = \frac{1}{40} p - \frac{1}{2}$
    - $q = \frac{1}{40} \frac{820}{9} - \frac{1}{2} = \frac{16}{9}$
    - $(\frac{16}{9}, \frac{820}{9})$
- Graph:
```
	left=-100; right=4000;
	top=160; bottom=-10;
	---
	x = 3600 - 20y | label: Demand
	x = 25y | label: Supply A (No Tax)
	x = 25y - 500 | label: Supply B (SR Tax)
	x = 20y - 400 | label: Supply C (LR Tax)
	(2000, 80) | label: A
	(1777.78, 91.11) | label: B
	(1600, 100) | label: C
```
```
	left=-0.5; right=3.5;
	top=150; bottom=-10;
	---
	y = 40x | label: MC
	y = 20x + 80/x | label: AC
	y = 40x + 20 | label: MC + Tax
	y = 20x + 20 + 80/x | label: AC + Tax
	(2, 80) | label: A
	(2, 100) | label: C
	(1.77777, 91.1111) | label: B
```
4
- $A$ and $B$
- $S M C = 2 q_{A}$
- $S M C = 4 q_{B}$
- $D = 60 - 0.75 p$
- Find market supply
  - $q_{A} = \frac{1}{2} p_{A}$
  - $q_{B} = \frac{1}{4} p_{B}$
- $S M C = q_{A} + q_{B}$
  - $S M C = \frac{1}{2} p + \frac{1}{4} p = \frac{3}{4} p$
  - $Q_{s} = n q$
  - We have two firms, so we can just add them up.
  - $Q_{s} = \frac{3}{4} p$
- $Q_{s_{A}} = D$
  - $\frac{1}{2} p = 60 - 0.75 p$
  - $p = 48 ⟹ q = 24$
- $Q_{s_{B}} = D$
  - $\frac{1}{4} p = 60 - 0.75 p$
  - $p = 60 ⟹ q = 15$
- $Q_{s} = D$
  - $\frac{3}{4} p = 60 - 0.75 p$
  - $p = 40.0 ⟹ q = 30$
- $(24, 48)$
- $(15, 60)$
- $(30, 40)$
```
    left=-2; right=70;
    top=90; bottom=-2;
    ---
    x=60-0.75p
    x=3/4 y
    x=1/4 y
    x=1/2 y
    (24,48)|Label:A
    (15,60)|Label:B
    (30,40)|Label:C
```

Week 12

Lecture

1
- $p = 60 - 2 q$
- $C = q^{2} + 100$
- $R = p q$
- $R = 60 q - 2 q^{2}$
- $M R = 60 - 4 q$
- $A C = q + 100 q^{- 1}$
- $M C = 2 q$
- $M R = M C$ , firms optimal
  - $60 - 4 q = 2 q$
  - $q = 10$
- $p = M C$ , socially optimal
  - $60 - 2 q = 2 q$
  - $q = 15$
- To subsidize to get to the socially optimal
- $M C - S = M R$
- $2 q - s = 60 - 4 q$
- $s = 6 q - 60 |_{q = 15} ⟹ s = 30$
- $s_{t} = 30 \cdot 15 = 450$

Tutorial

ECO101 - Week 12 Monopoly.pdf

1
- Monopolist
- $M C = Q$
- $p = 300 - Q$
- Eliminate efficiency loss with a subsidy
- $p = M C$
  - $p = Q$
  - $Q = 300 - Q$
  - $Q = 150 ⟹ p = 150$
  - $(150, 150)$
  - Socially optimal point
- $R = p q$
- $R = 300 q - q^{2}$
- $M R = 300 - 2 q$
- $M R = M C$
  - $q = 300 - 2 q$
  - $q = 100$
- $M C - s = M R$
- $q - s = 300 - 2 q$
- $s = 3 q - 300 |_{q = 150} ⟹ s = 150$
- $M C_{s} = q - 150$
```
    left=-2; right=305;
    top=305; bottom=-2;
    ---
    y=300-x
    y=x
    y=300-2x
    y=x-150
    (150,150)
    (100,100)
```
2
- Monopolist
- $p = 40 - q$
- $C = 0.5 q^{2}$
- $M C = Q$
- Calculate profits if they use a two-part tariff
- Socially optimal at $p = M C$
  - $40 - q = q$
  - $q = 20$
- Firm's optimal at $M R = M C$
  - $R = p q$
  - $R = (40 - q) q = - q^{2} + 40 q$
  - $M R = - 2 q + 40$
  - $M R = M C$
  - $- 2 q + 40 = q$
  - $q = \frac{40}{3}$
- Consumer surplus
  - $\frac{b h}{2} = \frac{(40 - 20) (20 - 0)}{2} = 200$
  - $200$ is our fee
- $π = R - C$
- $π = (- q^{2} + 40 q) - (\frac{1}{2} q^{2}) = \frac{q (80 - 3 q)}{2} = - \frac{3}{2} q^{2} + 40 q$
- $π = - \frac{3}{2} q^{2} + 40 q |_{q = 20}$
  - $- \frac{3}{2} (20)^{2} + 40 (20) = 200$
- $π = - \frac{3}{2} q^{2} + 40 q + 200 |_{q = 20}$
  - $= 400$
3
- $p_{1} = 14 - 0.5 q_{1}$ where $p_{1} = 10$
- $p_{2} = 10 - 0.5 q_{2}$
- $R = p q$
- $R = 10 q$ in both cases
- $M R = 10$
- Create $p = (14 - 0.5 q) + (10 - 0.5 q) = 24 - q$
- $1$
  - $10 = 14 - 0.5 q_{1}$
  - $q_{1} = 8.0$
  - $R = p q$
  - $R = (14 - \frac{1}{2} q_{1}) q_{1} = - \frac{q_{1}^{2}}{2} + 14 q_{1}$
  - $M R = - q_{1} + 14$
  - $M R (8) = - (8) + 14 = 6$
- $2$
  - $R = p q$
  - $R = 10 q - 0.5 q_{2}^{2}$
  - $M R = 10 - q_{2}$
- $M R 1 = M R 2$
  - $6 = 10 - q$
  - $q = 4$
  - $P = 10 - \frac{1}{2} (4) = 8$
```
    left=-2; right=30;
    top=15; bottom=-2;
    ---
    y=10-0.5x
    y=10-x
    y=10|dashed
    (8,6)
    (4,8)
```
```
    left=-2; right=30;
    top=15; bottom=-2;
    ---
    y=14-0.5x
    y=14-x
    (6,8)
```
4
- $c = 32 q - 0.05 q^{2}$
- $p = 40 - 0.15 q$
- $R = p q$
- $R = 40 q - 0.15 q^{2}$
- $M R = 40 - 0.3 q$
- $M C = 32 - 0.1 q$
- $M R = M C$
  - $40 - 0.3 q = 32 - 0.1 q$
  - $q = 40.0 ⟹ p = 28$
- $p = M C$
  - $40 - 0.15 q = 32 - 0.1 q$
  - $q = 160.0 ⟹ p = 16$
- Consumer Surplus
  - $p = 40 - 0.15 q |_{q = 0} ⟹ p = 40$
  - $\frac{b h}{2} = \frac{(40 - 16) (160 - 0)}{2} = 1920$
- Profits before
  - $π = R - C$
  - $π = (40 q - 0.15 q^{2}) - (32 q - 0.05 q^{2}) = - 0.1 q^{2} + 8 q$
  - $π = - 0.1 q^{2} + 8 q |_{q = 160} ⟹ - 1280$
- Profits after
  - $π = - 0.1 q^{2} + 8 q + 1920$
  - $π = - 1280 + 1920 = 640$
```
    left=-2; right=400;
    top=50; bottom=-2;
    ---
    y=40-0.15x | label: Demand (AR)
    y=40-0.3x | label: MR
    y=32-0.1x | label: MC
    (40,28) | label: MR=MC Intersection
    (40,34) | label: Monopoly Price
    (160,16) | label: Social Optimum
```

Final Exams

April 2023

eco101h-m23.pdf
1
- $S = 2 p - 40$
- $D = 120 - 2 p$
- Quota $Q = 20$
- $Q = 0$
  - $D = S$
    - $2 p - 40 = 120 - 2 p$
    - $p = 40 ⟹ q = 40$
  - PS
    - $q = 2 p - 40 |_{q = 0}$
    - $0 = 2 p - 40$
    - $p = 20$
    - $\frac{b h}{2} = \frac{(40 - 0) (40 - 20)}{2} = 400$
- $Q = 20$
  - $20 = 2 p - 40$
    - $p = 30 ⟹ q = 20$
  - $D (30) = 120 - 2 (30) = 60$
    - $(60, 30)$
    - $(20, 30)$
  - PS
    - $\frac{b h}{2} = \frac{(20 - 0) (30 - 20)}{2} = 100$
```
    left=-2; right=125;
    top=125; bottom=-2;
    ---
    x=2y-40
    x=120-2y
    y=30|dashed
    (40,40)
    (60,30)
    (20,30)
```
2
- Equilibrium prices
- $S = 2 p - 40$
- $D = 120 - 2 p + p_{y}$
- $p_{y}$ is a substitute in consumption
- $p_{y_{0}} = 40$
  - $D = S$
  - $120 - 2 p + 40 = 2 p - 40$
  - $p = 50 ⟹ q = 60$
- $p_{y_{1}} = 80$
  - $D = S$
  - $120 - 2 p + 80 = 2 p - 40$
  - $p = 60 ⟹ q = 80$
- $p_{y_{1}} |_{p = 50}$
  - $120 - (2) (50) + 80 = 100$
- Cross price elasticity of demand at original equilibrium
- $η = \frac{\frac{Δ Q}{\bar{Q}}}{\frac{Δ p}{\bar{p}}}$
- $η = \frac{\frac{40}{80}}{\frac{40}{60}} = \frac{3}{4}$
```
    left=-2; right=125;
    top=125; bottom=-2;
    ---
    x=2y-40
    x=120-2y+40
    x=120-2y+80
    (60,50)
    (80,60)
```
3
- 1. Roommate A with demand p = 100 – 20q and roommate B with demand p = 80 – 10q will purchase a non-rival, non-excludable public good at a price of 60 dollars per unit.
- a. Provide a Marginal Cost and Marginal Benefit Diagram to illustrate and quantify optimal quantities for this good.
- b. Roommate A should contribute $x$ dollars towards the purchase of this good.
- $A$ has $p = 100 - 20 q$
- $B$ has $p = 80 - 10 q$
- $M C = 60$ per unit
- $M B = M B_{A} + M B_{B}$
- $M B = (100 - 20 q) + (80 - 10 q) = 180 - 30 q$
- $M B = M C$
  - $180 - 30 q = 60$
  - $q = 4$ is socially optimal
- Split between
  - $M B_{A} = 100 - 20 (4) = 20$
  - $M B_{B} = 80 - 10 (4) = 40$
  - $A$ 's split is $(4) (\frac{20}{60}) = \frac{4}{3}$
  - $B$ 's split is $(4) (\frac{40}{60}) = \frac{8}{3}$
- So roommate $A$ should pay:
  - Total Cost
  - $4 \cdot 60 = 240$ total cost
  - $\frac{1}{3} (240) = 80$
4
- A utility maximizing consumer with $u (x, y) = \sqrt{x} \sqrt{y}$ pays $p_{x} = 4$ , $p_{y} = 1$ and has income $m = 320$ . If she joins a price club, she will pay $p_{x} = 1$ , $p_{y} = 1$ .
- a. Provide an Indifference Curve and Budget Line Diagram to illustrate and quantify the Substitution Effect and the Income Effect of this price change.
- b. The maximum amount she would pay to purchase a membership at this club is equal to $x$ dollars.
- $m = 320$
- $p_{x_{0}} = 4$
- $p_{y_{0}} = 1$
- $p_{x_{1}} = 1$
- $p_{y_{1}} = 1$
- Bundle A
  - Tangency
    - $M R S = O C O S T_{X}$
    - $\frac{y}{x} = \frac{p_{x_{0}}}{p_{y_{0}}}$
    - $4 x = y$
    - $x = \frac{1}{4} y$
  - Budget Line
    - $p_{x} x + p_{y} y = m$
    - $4 x + y = 320$
    - $4 x + (4 x) = 320$
    - $x = 40 ⟹ y = 160$
  - Utility
    - $u (x, y) = \sqrt{40} \sqrt{160} = 80$
- Bundle C
  - Tangency
    - $M R S = O C O S T_{X}$
    - $x = y$
  - Budget Line
    - $x + y = 320$
    - $2 x = 320$
    - $x = 160 ⟹ y = 160$
  - Utility
    - $u (x, y) = \sqrt{160} \sqrt{160} = 160$
- Price to pay
  - Utility of A
  - Prices of C
  - $u = 80$
  - $x + y = 320$
  - $u = \sqrt{x} \sqrt{y}$
  - Since $x = y$
  - $u = x$
  - $x = 80 ⟹ y = 80$
  - $80 + 80 = 160$
  - $320 - 160 = 160$ is the price to pay.
5
- A cost minimizing firm with production function $q = \sqrt{k} \sqrt{l}$ pays $w = 1$ , $r = 4$ and produces $q = 20$ units of output. Next month the firm will expand to $q = 40$ . In the short run the firm cannot change capital. In the long run both capital and labour can be varied.
- a. Provide an Isoquant and Isocost Diagram to illustrate the original choice as "A", the short run choice as "B" and long run choice as "C".
- b. The Short Run Total Cost of producing $q = 40$ is equal to $x$ .
- Cost minimizing firm
- $q = \sqrt{k} \sqrt{l}$
- $w = 1$
- $r = 4$
- $q_{0} = 20$
- $q_{1} = 40$
- Short run capital is fixed.
- Long run both are variable.
- Bundle A
  - Tangency
    - $M R T S = O C O S T$
    - $w l = r k$
    - $l = 4 k$
    - $k = \frac{1}{4} l$
  - Sufficiency
    - $q^{2} = k l$
    - $\frac{q^{2}}{l} = k$
    - $\frac{q^{2}}{k} = l$
  - Cost
    - $C = w l + r k$
    - $C = l + 4 k$
    - $C = l + l$
    - $C = 2 l$
  - Find point
    - $20^{2} = k l$
    - $20^{2} = 4 k^{2}$
    - $k = - 10 \lor k = 10 ⟹ l = 40$
    - $C = 2 (40) = 80$
  - Budget Line
    - $C = w l + r k$
    - $80 = l + 4 k$
    - $k = 20 - \frac{l}{4}$
- Bundle B
  - Short run, capital is the same, $k = 10$
  - Tangency
    - $M R T S = O C O S T$
    - $w l = r k$
    - $l = 4 k$
    - $k = \frac{1}{4} l$
  - Sufficiency
    - $q^{2} = k l$
    - $\frac{q^{2}}{10} = l$
  - Cost
    - $C = w l + r k$
    - $C = l + 4 k$
    - $C = l + (4) (10) = l + 40$
  - Find Points
    - $\frac{40^{2}}{10} = l$
    - $l = 160$
    - $k = 10$
    - $q = 40$
  - Budget Line
    - $C = l + 40$
    - $C = 160 + 40 = 200$
    - $200 = l + 4 k$
    - $k = 50 - \frac{l}{4}$
- Bundle C
  - Long run
  - Tangency
    - $M R T S = O C O S T$
    - $w l = r k$
    - $l = 4 k$
    - $k = \frac{1}{4} l$
  - Sufficiency
    - $q^{2} = k l$
    - $\frac{q^{2}}{k} = l$
    - $\frac{q^{2}}{l} = k$
  - Cost
    - $C = w l + r k$
    - $C = l + 4 k$
    - $C = l + (4) (\frac{q^{2}}{l}) = l + 4 \frac{q^{2}}{l}$
    - $C = l + 4 \frac{q^{2}}{l}$
    - $C = l + 4 \frac{(40)^{2}}{l} = l + 6400 \frac{1}{l}$
  - Find Points
    - $\frac{40^{2}}{l} = k$
    - $1600 = k l$
    - $1600 = 4 k^{2}$
    - $k = - 20 \lor k = 20 ⟹ l = 80$
  - Budget Line
    - $C = 80 + \frac{6400}{80} = 160$
    - $160 = l + 4 k$
    - $k = 40 - \frac{l}{4}$
```
    left=-2; right=330;
    top=90; bottom=-2;
    ---
    y=40-x/4
    y=50-x/4
    y=20-x/4
    (80,20)
    (160,10)
    (40,10)
    1600=xy
    400=xy
```
6
- Each firm in a constant costs competitive industry has C = 2q2 + 8 and MC = 4q. Demand is Q = 480 – 20P. Next month the government will introduce a lump sum tax of 24 dollars. a. Provide the usual 2-panel diagram showing "The Firm" and "The Industry" to illustrate and quantify the impact of this tax. b. Total tax revenues are equal to x dollars.
- $C = 2 q^{2} + 8$
- $A C = 2 q + 8 q^{- 1}$
- $M C = 4 q$
- $Q = 480 - 20 P$
- Lump Sum Tax of $24$
- Bundle A
  - $M C = A C$
    - $4 q = 2 q + 8 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 4 (2) = 8$
  - $(2, 8)$
  - Firm costs
    - $C = (2) (2)^{2} + 8 = 16$
  - $Q$
    - $480 - 20 (8) = 320$
  - $n$
    - $n = \frac{320}{2} = 160$
- Bundle C
  - $C = 2 q^{2} + 8 + 24 = 2 q^{2} + 32$
  - $M C = 4 q$
  - $A C = 2 q + 32 q^{- 1}$
  - $M C = A C$
    - $4 q = 2 q + 32 q^{- 1}$
    - $q = - 4 \lor q = 4$
  - $p = M C$
    - $p = 4 (4) = 16$
  - $(4, 16)$
  - Firm Costs
    - $C = (2) (4)^{2} + 32 = 64$
  - $Q$
    - $480 - 20 (16) = 160$
  - $n$
    - $n = \frac{160}{4} = 40$
- Graphs:
```
    left=-2; right=500;
    top=50; bottom=-2;
    ---
    x=480-20y
    y=16|dashed
    y=8|dashed
    (160,16)
    (320,8)
```
```
    left=-2; right=10;
    top=20; bottom=-2;
    ---
    y=4x
    y=16|dashed
    y=8|dashed
    (4,16)
    (2,8)
	y=(x-4)^2+16
	y=(x-2)^2+8|x>0
```
- Tax revenue
  - $t \cdot n$
  - $24 \cdot 40 = 960$

ECO101 June 2023 Exam

Perfectly competitive constant cost market
Long run equilibrium is $p = 1$
The government requires each seller must have a $500$ dollar license
Because you increase the costs, prices will go up and demand will go down.

Week 11 Again

Tutorial

ECO101 - Week 11 Competition.pdf

1
- $D = 680 - 20 p$
- Per firm:
- $\begin{array}{l} Output & 1 & 2 & 3 & 4 & 5 & 6 \\ Cost & 8 & 9 & 12 & 17 & 24 & 33 \\ MC & 0 & 2 & 4 & 6 & 8 & 10 \\ AC & 8 & 4.5 & 4 & 4.25 & 4.8 & 5.5 \end{array}$
- $M C = A C$ is at $q = 3$
- $p = M C$
- $p = 4$
- $Q = 680 - 20 (4) = 600$
- $n = \frac{Q}{q}$
- $n = \frac{600}{3} = 200$
- There will be $200$ firms
2
- Firms $C = 20 q^{2} + 80$
- $M C = 40 q$
- $A C = 20 q + 80 q^{- 1}$
- Market Demand is $Q = 2400 - 20 p$
- $S = 20$ subsidy
- Bundle A: $S = 0$
  - $M C = A C$
    - $40 q = 20 q + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 (2) = 80$
  - $Q = 2400 - 20 (80) = 800$
  - $n = \frac{Q}{q}$
    - $n = \frac{800}{2} = 400$
  - $Q_{s} = n p$
    - $p = M C$
    - $p = 40 q$
    - $\frac{1}{40} p = q$
    - $\frac{n}{40} p$
    - $\frac{400}{40} p = 10 p$
- Bundle C: $S = 20$
  - Subsidy reduces costs
  - $C = 20 q^{2} + 80 - 20 q$
  - $M C = 40 q - 20$
  - $A C = 20 q - 20 + 80 q^{- 1}$
  - $M C = A C$
    - $40 q - 20 = 20 q - 20 + 80 q^{- 1}$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 q - 20$
    - $p = 40 (2) - 20 = 60$
  - $Q = 2400 - 20 (60) = 1200$
  - $n = \frac{Q}{q}$
    - $n = \frac{1200}{2} = 600$
  - $Q_{s} = n p$
    - $p = M C$
    - $p = 40 q - 20$
    - $p + 20 = 40 q$
    - $\frac{p + 20}{40} = q$
    - $n \frac{p + 20}{40}$
    - $600 \frac{p + 20}{40} = 15 p + 300$
- Bundle B: Same firm amount as A with subsidy
  - $C = 20 q^{2} + 80 - 20 q$
  - $M C = 40 q - 20$
  - $A C = 20 q - 20 + 80 q^{- 1}$
  - $n = 400$
  - $M C = A C ⟹ q = 2, p = 60$
  - $n = \frac{Q}{q}$
  - $400 = \frac{Q}{q}$
  - $400 = \frac{Q}{2}$
  - $Q = 800$
  - $Q_{s} = n p$
    - $p = M C$
    - $p = 40 q - 20$
    - $q = \frac{p}{40} + \frac{1}{2}$
    - $Q_{s} = 400 (\frac{p}{40} + \frac{1}{2}) = 10 p + 200$
    - $10 p + 200 = 2400 - 20 p$
    - $p = \frac{220}{3}$
    - $10 (\frac{220}{3}) + 200 = \frac{2800}{3}$
  - $\frac{220}{3} = 40 q - 20$
  - $q = \frac{7}{3}$
- Graph:
  - Industry:
```
    left=-2; right=2500;
    top=150; bottom=-2;
    ---
    x=2400-20y
    x=15y+300
    x=10y+200
    x=10y
    (800,80)
    (1200,60)
    (2800/3,220/3)
    y=80|dashed
    y=60|dashed
    y=220/3|dashed
```
  - Firm:
```
    left=-2; right=20;
    top=150; bottom=-2;
    ---
    y=40x-20
    y=40x
    (2,80)
    (2,60)
    (7/3,220/3)
    y=80|dashed
    y=60|dashed
    y=220/3|dashed
    y=(x-2)^2+80
    y=(x-2)^2+60
    y=(x-7/3)^2+220/3
```
3
- $C = 20 q^{2} + 80$
- $M C = 40 q$
- $Q = 3600 - 20 p$
- $t = 20$ tax
- Bundle A: $t = 0$
  - $A C = 20 q + 80 q^{- 1}$
  - $A C = M C$
    - $20 q + 80 q^{- 1} = 40 q$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 q$
    - $p = 40 (2) = 80$
  - $Q = 3600 - 20 (80) = 2000$
  - $n = \frac{Q}{q}$
    - $n = \frac{Q}{q}$
    - $n = \frac{2000}{2} = 1000$
    - Firms
  - $Q_{s} = n p$
    - $p = M C$
    - $p = 40 q$
    - $q = \frac{p}{40}$
    - $Q_{s} = 1000 (\frac{p}{40}) = 25 p$
- Bundle C: $t = 20$
  - Tax makes everything cost more
  - $C = 20 q^{2} + 80 + 20 q$
  - $M C = 40 q + 20$
  - $A C = 20 q + 80 q^{- 1} + 20$
  - $Q = 3600 - 20 p$
  - $M C = A C$
    - $40 q + 20 = 20 q + 80 q^{- 1} + 20$
    - $q = - 2 \lor q = 2$
  - $p = M C$
    - $p = 40 q + 20$
    - $p = 40 (2) + 20 = 100$
    - $p = M C$
    - $p = 40 q + 20$
    - $q = \frac{p}{40} - \frac{1}{2}$
  - $Q = 3600 - 20 p$
    - $Q = 3600 - 20 (100) = 1600$
  - $n = \frac{Q}{q}$
    - $n = \frac{1600}{2} = 800$
  - $Q_{s} = n p$
    - $800 (\frac{p}{40} - \frac{1}{2}) = 20 p - 400$
- Bundle B: $t = 20$ , $n = 1000$
  - Firms haven't had time to exit yet.
  - $C = 20 q^{2} + 80 + 20 q$
  - $M C = 40 q + 20$
  - $A C = 20 q + 80 q^{- 1} + 20$
  - $Q = 3600 - 20 p$
  - $n = 1000$
  - $M C = A C ⟹ q = 2, p = 100$
  - $n = \frac{Q}{q}$
  - $1000 = \frac{Q}{2}$
  - $Q = 2000$
  - $p = M C$
    - $p = 40 q + 20$
    - $q = \frac{p}{40} - \frac{1}{2}$
  - $Q_{s} = n p$
    - $Q_{s} = (1000) (\frac{p}{40} - \frac{1}{2}) = 25 p - 500$
    - $3600 - 20 p = 25 p - 500$
    - $p = \frac{820}{9}$
    - $q = 3600 - 20 (\frac{820}{9}) = \frac{16000}{9}$
  - $40 q + 20 = \frac{820}{9}$
  - $q = \frac{16}{9}$
- Graph
- Industry
```
    left=-2; top=190;
    right=3610; bottom=-2;
    ---
    x=3600-20y
    x=25y-500
    x=20y-400
    x=25y
    (1600,100)
    (2000,80)
    (16000/9,820/9)
    y=100|dashed
    y=80|dashed
    y=820/9|dashed
```
- Firm
```
    left=-2; top=200;
    right=10; bottom=-2;
    ---
    y=40x
    y=40x+20
    (2,100)
    (2,80)
    (16/9,820/9)
    y=100|dashed
    y=80|dashed
    y=820/9|dashed
    y=4(x-2)^2+100
    y=4(x-2)^2+80
```
4
- $S M C = 2 q_{A}$
- $S M C = 4 q_{B}$
- $D = 60 - 0.75 p$
- Find the market supply
- $p = M C$
- $p = 2 q_{A}$
  - $\frac{1}{2} p = q_{A}$
- $p = 4 q_{B}$
  - $\frac{1}{4} p = q_{B}$
- $q_{A} = 2 q_{B}$
- Aggregate
  - $Q_{s} = \frac{3}{4} p$ because there's two firms
- Solve for equilibrium price and quantity
  - $Q_{s} = Q_{d}$
  - $\frac{3}{4} p = 60 - 0.75 p$
  - $p = 40.0 ⟹ Q = 30$
  - $p = M C$
    - $p = 2 q_{A}$
    - $40 = 2 q_{A}$
    - $q_{A} = 20$
  - $p = M C$
    - $p = 4 q_{B}$
    - $40 = 4 q_{B}$
    - $q_{B} = 10$
  - $(20, 40)$
  - $(10, 40)$
  - $(30, 40)$
- Graphs:
- Industry
```
    left=-2; right=70;
    top=90; bottom=-2;
    ---
    x=60-0.75y
    x=3/4y
    (30,40)
    y=40|dashed
```
- Firm
```
    left=-2; right=70;
    top=90; bottom=-2;
    ---
    x=3/4y
    x=1/4y
    x=1/2y
    (20,40)
    (10,40)
    (30,40)
    y=40|dashed
```
$0.4 x + 0.4 (\frac{2}{3}) + 0.2 (\frac{281.21}{320}) = \frac{63}{100}$
$x = 0.468942708333333$