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Homework Assignment #5
14.27 Consider the production function Q = K 3/4 L3/4 . Show that marginal productivity of each factor is
diminishing. Show, however, that for any strictly positive input combination, if the input combination
is doubled, then output more than doubles.
Answer: The marginal products are MPK = (3/4)K −1/4 L3/4 and MPL = (3/4)K 3/4 L−1/4 , which are
decreasing in K and L, respectively. However, Q(2K, 2L) = 23/2 K 3/4 L3/4 = 23/2 Q(K, L) > 2Q(K, L).
15.6 Consider the function F (x1 , x2 , y) = x21 − x22 + y 3 .
a) If x1 = 6 and x2 = 3, find a y which satisfies F (x1 , x2 , y) = 0.
Answer: The equation becomes 27 + y 3 = 0, which is satisfied by y = −3 (this is the only real
solution, but there are also two complex solutions, y = 3e±2i/3 ).
b) Does this equation define y as an implicit function of x1 and x2 near x1 = 6, x2 = 3?
Answer: We calculate ∂F/∂y(1, 3, −3) = 3y 2 (1, 3, −3) = 3 · (−3)2 = 27 6= 0. By the Implicit
Function Theorem, y can be implicitly defined as a function of (x1 , x2 ) in a neighborhood of (1, 3).
c) If so, compute ∂y/∂x1 (6, 3) and ∂y/∂x2 (6, 3).
∂y
∂F ∂F
(6, 3, −3) = −( ∂x
/ ∂y )(6, 3, −3) = −(2x1 /3y 2 )(1, 3, −3) = −4/9 and
Answer: Here ∂x
1
1
∂F ∂F
−( ∂x
/ ∂y )(6, 3, −3) = (2x2 /3y 2 )(6, 3, −3) = 6/27 = 2/9.
2
∂y
∂x2
=
d) If x1 increases to 6.2 and x2 decreases to 2.9, estimate the corresponding change in y.
Answer: The change is ∆y ≈ −(4/9)∆x1 + (2/9)∆x2 = −(4/9)(.2) − (2/9)(.1) = −1/9 ≈ −0.11.
15.19 Does the system xz 3 + y 2 v 4 = 2, xz + yvz 2 = 2 define v and z as C 1 functions of x and y around the
point (1, 1, 1, 1)? If so, find ∂z/∂x, ∂z/∂y, ∂v/∂x, and ∂v/∂y there.
3
xz + y 2 v 4 − 2
. We calculate
Answer: Let F =
xz + yvz 2 − 2
dv,z F (1, 1, 1, 1) =
4y 2 v 3
yz 2
4
3xz 2
=
1
x + 2yvz (1,1,1,1)
3
.
3
3 −3
Its determinant is 9 6= 0, so dv,z F (1, 1, 1, 1) is invertible. In fact, the inverse is
.
−1 4
The Implicit Function Theorem then tells us that v and z can be written as C 1 functions of (x, y)
near (x, y) = (1, 1), and that the derivative matrix at (1, 1) is:
1
9
∂v/∂x
∂z/∂x
3
∂v/∂y
z 2yv 4 1 3 −3
(1, 1) = −
×
9 −1 4
∂z/∂y
z
vz 2 (1,1,1,1)
1 2
1 3 −3
=−
×
9 −1 4
1 1
1 0 3
=−
.
9 3 2
15.36 Show that the map F (x, y) = (x + ey , y + e−x ) is everywhere locally invertible.
Answer: Assuming F is a column vector, we compute
dF =
1
e−x
−ey
.
1
HOMEWORK ASSIGNMENT #5
Page 2
y−x
Its determinant is 1 + e
> 1 > 0. Since dF is always invertible, the Inverse Function Theorem
guarantees the existence of a local inverse about any point (x0 , y0 ).
16.6 Determine the definiteness of the following constrained quadratics:
a) Q(x1 , x2 ) = x21 + 2x1 x2 − x22 , subject to x1 + x2 = 0.
1 1
Answer: Here A =
and the bordered Hessian is
1 −1
0
H = 1
1

1
1
1

1
1 
−1.
Here n = 2 and m = 1, so the only leading principal minor we need check is H3 = +2. As it is
positive (same sign as (−1)n ), the constrained quadratic is negative definite.
b) Q(x1 , x2 ) = 4x21 + 2x1 x2 − x22 , subject to x1 + x2 = 0.
4 1
Answer: Here A =
. The Hessian is
1 −1
0

H= 1
1

1
4
1

1
1 .
−1
Here n = 2 and m = 1, so the only leading principal minor we need check is H3 = −1. As it is
negative (same sign as (−1)m ), the constrained quadratic is positive definite.
c) Q(x1 , x2 , x3 ) = x21 + x22 − x23 + 4x1 x3 − 2x1 x2 , subject to x1 + x2 + x3 = 0 and x1 + x2 − x3 = 0.
Answer: Here

1
A =  −1
2

−1
1
0
0


2
0

0  and H =  1

1
−1
1
0
0
1
1
−1
1
1
1
−1
2

1
−1 


2 .

0 
−1
1
1
−1
1
0
Here n = 3 and m = 2, so the only leading principal minor we need check is H5 = 16. This has the
same sign as (−1)m , so the constrained quadratic is positive definite.
d) Q(x1 , x2 , x3 ) = x21 + x22 + x23 + 4x1 x3 − 2x1 x2 , subject to x1 + x2 + x3 = 0 and x1 + x2 − x3 = 0.
Answer: Here

1
A =  −1
2

−1
1
0
0

2
0

0  and H =  1

1
1
1

0
0
1
1
−1
1
1
1
−1
2
1
1
−1
1
0

1
−1 


2 .

0 
1
Here n = 3 and m = 2, so the only leading principal minor we need check is H5 = 16. This has the
same sign as (−1)m , so the constrained quadratic is positive definite.
e) Q(x1 , x2 , x3 ) = x21 − x23 + 4x1 x2 − 6x2 x3 , subject to x1 + x2 − x3 = 0.
HOMEWORK ASSIGNMENT #5
Page 3
Answer: Here
1
A = 2
0

2
0
−3

0

0
 1
−3  and H = 
 1
−1
−1
1 1
1 2
2 0
0 −3

−1
0 
.
−3 
−1
Here n = 3 and m = 1, so we must check the last 2 leading principal minors H3 = 3 and H4 = 4.
Then H4 has a different sign from (−1)n = −1 and (−1)m = −1. The minors are non-zero and
violate the pattern, so the constrained form is indefinite.

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