Answers to Second Test, ECMA04H, November 7, 2003
1.
If q = 10L
1/3
, then the marginal product of labour is dq/dL = 10 x 1/3 L
-2/3
.
At L = 8, this is
10/12 = 0.833 or 5/6.
The correct answer is (C).
2.
The marginal cost per unit of output is given by the price of an extra unit of labour divided by
the marginal product of that labour.
In this case, this is 4/(10/3 L
-2/3
), evaluated at L = 125.
This is 4/(10/75), which is $30.
The correct answer is (O).
3.
The total cost function is given by the price of capital times the number of units of capital
employed, plus the price of labour times the number of units of labour.
This is (27 x 8) + (4 x
L).
The average cost function is TC/q, where q = 10L
1/3
, so AC = 216/(10L
1/3
) + (4L)/
(10L
1/3
).
This can be written as 21.6 L
-1/3
+ 0.4L
2/3
.
Taking the derivative of AC with respect
to L, we have dAC/dL = -7.2L
-4/3
+ 4/15 L
-1/3
.
Setting this derivative equal to zero, we have
L
4/3
/ L
1/3
= (15 x 7.2)/4 or L = 27.
The correct answer is (M).
4.
TC = 2q
2
+ 100q + 200 and FC = 72.
We know that TC = VC + FC, so AC = AVC + AFC, or
AFC = AC – AVC.
When q = 144, AFC = 72/144 = $0.50.
In other words, the difference
between AC and AVC is $0.50.
The correct answer is (D).
5.
TC = 2q
2
+ 100q + 200 so AC = 2q + 100 + 200q
-1
.
dAC/dq = 2 – 200q
-2
, which we can set
equal to zero to find the minimum.
If 2 – 200q
-2
= 0, then q
2
= 100, or q = 10.
At q = 10, AC
= (2 x 10) + 100 + 200/10 = $140.
The correct answer is (O).
6.
If P = 128, firms will profit maximize by setting P = MC.
MC = 4q + 100, so 128 = 4q +
100, or q = 7.
At this output, profit can be found as TR – TC = (128 x 7) – [(2 x 7 x 7) + (100
x 7) + 200] = 896 – 998 = -$102.
This is the best the firm could do while continuing to