Homework 1 - Math 105, Section 204
Due at the beginning of lecture on January 20, 2012
Name:
SID:
1. Consider the production function
3
1
f (x, y) = 60x 4 y 4 ,
which gives the number of units of goods produced when utilizing x units
of labor and y units of capital.
and ∂f
. These quantities are referred to as the marginal
(a) Compute ∂f
∂x
∂y
productivities of labor and of capital respectively.
Solution: We have
∂f
3 −1/4 1/4
y = 45x−1/4 y 1/4 ,
= 60
x
∂x
4
and
∂f
1 −3/4
3/4
= 60x
y
= 15x3/4 y −3/4 .
∂x
4
(b) A firm with this production function currently operates with 16 units
of capital and 81 units of labor. If the amount of capital is held fixed
at y = 16 and the amount of labor increases by 1 unit, estimate the
increase in the quantity of goods produced.
Solution: The formula for estimating change is given by
∆f ≈ f( x0 , y0 )∆x + fy (x0 , y0 )∆y.
The problem gives
x0 = 81, y0 = 16, ∆x = 1, ∆y = 0.
Therefore ∆f is approximately
161/4
2
45 1/4 = 45 = 30.
3
81
(c) Suppose instead that the amount of labor is held fixed at x = 81 and
the amount of capital increases by 1 unit. Estimate the increase in the
quantity of goods produced.
Solution: Using the same formula as before, we have
x0 = 81, y0 = 16, ∆x = 0, ∆y = 1.
Therefore ∆f is approximately
15
813/4
27
405
= 50.625.
= 15 =
3/4
8
8
16
1
2
2. A monopolist markets a product in two countries and can charge different
amounts in each country. Let x be the number of units to be sold in the
first country and y the number of units to be sold in the second country.
Due to the laws of demand, the monopolist must set the price at
97 − (x/10) dollars in the first country and 83 − (y/20) dollars in the
second country to sell all units. The cost of producing these units is
20, 000 + 3(x + y). Find the values of x and y that maximize the profit.
Solution: The revenue is given by
x
y
x 97 −
+ y 83 −
,
10
30
while the cost is
20000 + 3(x + y).
Therefore the profit function is
x
y
x2 y 2
x 97 −
+y 83 −
−[20000 + 3(x + y)] = 94x+80y − − −20000.
10
30
10 20
To find the maximum, we locate the critical points. The partial
derivatives are
x
∂f
= 94 − ,
∂x
5
∂f
y
= 80 − ,
∂y
10
so the only critical point is x = 470, y = 800. We use the second derivative
test to check what type of critical point it is. The second partial
derivatives are
1
∂ ∂f
=− ,
∂x ∂x
5
∂ ∂f
1
=− ,
∂y ∂y
10
∂ ∂f
∂ ∂f
=
= 0.
∂x ∂y
∂y ∂x
We have
1
D = fxx fyy − fxy fyx =
> 0,
50
and fx x < 0, so the critical point is a maximum. Since it’s the only one,
it’s also a global maximum, so x = 470, y = 800 maximizes the profit.
(Note: we do not need to check if any maximums occur on the boundary
because our work above holds for all x, y.)
3
3. Does there exist a function F (x, y) such that Fx (x, y) = xy, Fy (x, y) = y 2 ?
If yes, find such a function. If not, explain why not.
Solution: By Clairaut’s Theorem, if G is a function with second partial
derivatives, and Gxy , Gyx are both continuous on a region D, then
Gxy = Gyx
for any point in D. If F exists, then we have Fxy = x, Fyx = 0, and both
of these are continuous over the real numbers, but not equal. This
contradicts Clairaut’s Theorem, so F can’t exist.