Cun: 0993718306
Assignment 2
Question 1.
𝑇𝑃𝑃 = 0.4𝐿 + 0.09𝐿2 − 0.0035𝐿3 ;  𝑀𝑃𝑃 = 0.4 + 0.18𝐿 − 0.0105𝐿2 ;
a. 𝐴𝑃𝑃 =
𝑇𝑃𝑃
𝐿
= 0.4 + 0.09𝐿 − 0.0035𝐿2
b. 𝑇𝑃𝑃 is maximized when MPP is equal to zero 
0.4 + 0.18𝐿 − 0.0105𝐿2 = 0
−0.18 ± √0.182 − 4(−0.0105)(0.4)
2(−0.0105)
You got two values
𝐿=
𝐿 = 19.13 (𝑡ℎ𝑜𝑢𝑠 ℎ𝑜𝑢𝑟𝑠)𝑜𝑟 𝐿 = −1.99 (𝑡ℎ𝑜𝑢𝑠 ℎ𝑜𝑢𝑟𝑠)
 L* =1 9.130 (thous. hours) will maximize TPP (why not the other L?).
c. APP is maximized when APP = MPP 
0.4 + 0.09𝐿 − 0.0035𝐿2 = 0.4 + 0.18𝐿 − 0.0105𝐿2
0.09𝐿 − 0.0035𝐿2 = 0.18𝐿 − 0.0105𝐿2
0.09𝐿 − 0.007𝐿2 = 0
𝐿(0.09 − 0.007𝐿) = 0
𝐿 = 0 ; 0.09 − 0.007𝐿 = 0 → 𝐿 = 12.857 (𝑡ℎ𝑜𝑢𝑠. ℎ𝑜𝑢𝑟𝑠)
 L* =12.857 (thous hours) will maximize MPP (why not the other L?).
At this level of labor, the output is:
𝑇𝑃𝑃 = 0.4𝐿 + 0.09𝐿2 − 0.0035𝐿3 = 0.4 ∗ 12.857 + 0.09 ∗ 12.8572 − 0.0035 ∗ 12.8573 =12.581
d. Stage 1: 0≤L<12.857 (thous hours)
Stage2: 12.857≤L≤19.13 (thous hours)
Stage 3: L>19.13 (thous hours)
e.
Total Physical Product
Output
19.0
14.0
9.0
4.0
(1.0)
3
5
7
9 11 13 15 17 19 21 23 25 27 29 31 33 35
(6.0)
Lab ('000
hours)
(11.0)
(16.0)
(21.0)
(26.0)
APP and MPP
Output
1.50
0.50
(0.50) 0
10
20
30
Lab ('000 hours)
(1.50)
APP
(2.50)
(3.50)
(4.50)
(5.50)
(6.50)
f) Poutput = 10; Pinput = 11
Profit-maximizing labor use:
MPP
The firm maximizes profit  MVP = MIC
MPP * Pout put = Pinput
MPP = Pinput / Poutput = 11/10 = 1.1
0.4 + 0.18𝐿 − 0.0105𝐿2 = 1.1
−0.7 + 0.18𝐿 − 0.0105𝐿2 = 0
You got two values:
𝐿 = 11.18 (𝑡ℎ𝑜𝑢𝑠 ℎ𝑜𝑢𝑟𝑠)𝑜𝑟 𝐿 = 5.96 (𝑡ℎ𝑜𝑢𝑠 ℎ𝑜𝑢𝑟𝑠)
You notice that both levels of labor use we have found are in stage 1. So if firm wants to maximize
profit, it should not produce at this stage.
Bonus
MPP is maximized when its slope equals, or ∂MPP/∂L , equals to zero:
∂𝑀𝑃𝑃/ ∂𝐿 = 0.18 − 0.021𝐿 = 0
0.18 = 0.021𝐿
 LMPP max = 8.57 (thous hrs)
Question 2.
Table 1. Use of inputs X1, X2 , and output:
X1
100
200
300
400
500
100
1000
1400
1600
1750
1850
200
1400
2000
2400
2600
2750
X2
300
1600
2400
3000
3400
3550
400
1750
2600
3400
4000
4300
500
1850
2750
3550
4300
5000
a) X1 fixed, X2 varies:
Using MPP = Q /X2
Apply this for each level of X1 (fixed), you will have Table 2 below.
For example at X1 = 200: When X2 increases from 200 to 300, output increases from 2000 to 2400 
MPPX2 = (2400-2000)/(300-200) = 4
Table 2. Marginal Physical Product of X2 at different levels of X1
MPP
X1=300
X2
X1=100 X1=200
X1=400 X1=500
100
200
4
6
8
8.5
9
300
2
4
6
8
8
400
1.5
2
4
6
7.5
500
1
1.5
1.5
3
7
The table shows that, at all levels of X1, we observe the evidence of the law of diminishing marginal
returns (or MPPX2 is decreasing).
For example, at Q1 = 200, MPPX2 decreases from 6, to 4, then 2, then 1.5 when X2 increases from 100, to
200, 300, 400, and 500.
b) X2 fixed, X1 varies
Notice that the cells inside Table 1 forms a symmetric matrix for X1 and X2  Every result calculated for
X2 applies for X1  Without further calculation, you will have Table 3:
Table 3. Marginal Physical Product of X1 at different levels of X2
X1
X2=100 X2=200
100
200
4
6
300
2
4
400
1.5
2
500
1
1.5
MPP
X2=300
X2=400
8
6
4
1.5
8.5
8
6
3
X2=500
9
8
7.5
7
Question 3.
𝑇𝐶 = 5 + 10𝑄 − 0.9𝑄 2 + 0.04𝑄 3
𝑀𝐶 = 10 − 1.8𝑄 + 0.12𝑄 2
a) TFC = 5; AFC = TFC/Q = 5/Q
b) 𝑇𝑉𝐶 = 10𝑄 − 0.9𝑄 2 + 0.04𝑄 3 ; 𝐴𝑉𝐶 = 10 − 0.9𝑄 + 0.04𝑄 2 ;
5
𝑇𝐶
𝐴𝑇𝐶 = 𝐴𝐹𝐶 + 𝐴𝑉𝐶 = + 10 − 0.9𝑄 + 0.04𝑄 2 (𝑜𝑟 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑤𝑎𝑦: 𝐴𝑇𝐶 =
)
𝑄
𝑄
c)
MC, AFC, & AVC
Cost
200
180
160
140
120
MC
100
AFC
80
AVC
60
40
20
Output (0'000')
0
0.1
5.1
10.1
15.1
20.1
25.1
30.1
35.1
d) AVC is minimized when AVC = MC
10 − 0.9𝑄 + 0.04𝑄 2 = 10 − 1.8𝑄 + 0.12𝑄2
−0.9𝑄 + 0.04𝑄 2 = −1.8𝑄 + 0.12𝑄 2
−0.9𝑄 + 0.08𝑄 2 = 0
 𝑄 = 0; 𝑄 = 11.25
So AVC is minimized when QAVC = 112,500 (loaves)
e) Profit is maximized when P = MR = MC  4 = MC
10 − 1.8𝑄 + 0.12𝑄 2 = 4
6 − 1.8𝑄 + 0.12𝑄 2 = 0
𝑄=
[1.8 ± √1.82 − 4(6)(0.12)]
(2)(0.12)
𝑄 = 5 𝑜𝑟 𝑄 = 10
Both of the two values that maximizes profit at price of 4 are smaller than the value ofQ AVC 
shut down firm makes a loss of 5 for the whole fixed cost.