Download Homework Assignment #6

Homework Assignment #6
17.4 A firm uses two inputs to produce a single product. If its production function is Q = x1/4 y 1/4 and if it
sells its output for a dollar a unit and buys each input for $4 dollars a unit, find its profit-maximizing
input bundle. (Check the second order conditions.)
Answer: Under these conditions, profit is π(x, y) = x1/4 y 1/4 − 4x − 4y. We treat this as a unconstrained profit since neither x = 0 nor y = 0 can yield positive profit. The first order conditions are
(1/4)x−3/4 y 1/4 = 4 and (1/4)x1/4 y −3/4 = 4. Dividing the first by the second yields y/x = 1, so x = y.
Then substitute back in the first equation to find (1/4)x−1/2 = 4. The solution is x = y = 1/256.
The Hessian of the objective function is:
H=
1
16
−3x−7/4 y 1/4
x−3/4 y 1/4
x−3/4 y 1/4
−3x1/4 y −7/4
.
The leading principal minors are H1 = −(3/16)x−7/4 y 1/4 < 0 and H2 = (1/256)(8x−3/2 y −3/2 ) > 0,
indicating that H is negative definite. This means (x, y) = (1/256, 1/256) is a maximum.
17.6 Dingbat Airlines has regular flights between Ypsilanti and Kalamazoo. It can treat business and pleasure
travelers as separate markets by demanding advance purchase and Saturday night stay-over for pleasure
travelers. Supose that it notes a demand function of Q = 16 − p for business travelers and a demand
function Q = 10−p for pleasure travelers and that it has a cost function for all travelers of C(Q) = 10+Q2 .
How much should it charge in each market to maximize its profit?
Answer: Let pb and pp be the prices in each market. Then revenue is pp (10 − pp ) + pb (16 − pb ), the
quantity produced is 26 − pb − pb and cost is 10 + (26 − pb − pp )2 . Profit is π(pb , pp ) = pp (10 − pp ) +
pb (16 − pb ) − 10 − (26 − pb − pp )2 . The first order conditions are
∂π
= 68 − 4pb − 2pp = 0 and
∂pb
∂π
= 62 − 2pb − 4pp = 0.
∂pp
The Hessian is:
H=
−4
−2
−2
−4
.
The leading principal minors are H1 = −4 and H2 = 12, indicating that we have a maximium. The
solution is pb = 37/3 and pp = 28/3.
17.9
a) Prove that 2ab ≤ a2 + b2 for all numbers a, b.
Answer: Use 0 ≤ (a − b)2 = a2 − 2ab + b2 . Rearrange to complete the proof.
b) Use this result to show that
(x1 + · · · + xn )2 = x21 + · · · + xn +
X
2xi xj
i<j
≤ x21 + · · · + x2n + (n − 1)(x21 + · · · + x2n )
= n(x21 + · · · + x2n )
HOMEWORK ASSIGNMENT #6
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(x2i
Answer: The first equality comes from carrying out the multiplication. Now 2xi xj ≤
+ x2j ).
P
P
When we look at the sum i<j 2xi xj ≤ i<j x2i + x2j , we find that each xk appears (n − 1) times,
P
Pn
so i<j x2i + x2j = (n − 1) i=1 x2i , yielding line 2. Line follows immediately.
c) Conclude that the point (m∗ , b∗ ) in (14) and (15) is a global minimizer of the function S in (11).
Answer: The Hessian of S is
H=
2
2n
P
xi
P
i
P
2 i xi
P 2 .
2 i xi
P
Here H1 = 2 > 0 and H2 = 4n[( i x2i ) − ( i xi )2 ]. Unless xi = xj = x for all i, j, H2 > 0. Note
that in that case, the other first-order principal minor is non-negative. It follows that H is positive
semidefinite for all x ∈ Rn (and mostly positive definite). Therefore the solution to the first order
conditions is a minimum.
18.2 Find the maximum and minimum distance from the origin to the ellipse x2 + xy + y 2 = 3. [Hint: Use
x2 + y 2 as your objective function.]
Answer: The Lagrangian is L = x2 +y 2 −µ(x2 +xy +y 2 −3) with first order conditions 2x−2µx−y = 0
and 2y − 2µy − x = 0. The solutions to the first order conditions are µ = 1, x = 0, y = 0, and µ 6= 1 with
x2 = y 2 . The first set fails to obey the constraint, so either 3x2 = 3 or x2 = 3, depending on whether
x = y or x = −y. The minimum distance is at (1, 1) and (−1, −1) and the maximum distance is at
√
√
√ √
( 3, − 3) and (− 3, 3). We know these are maxima and minima by computing the distance at each
point.
18.12 Consider the problem of maximizing f (x, y, z) = xyz + z, subject to the constraints x2 + y 2 + z ≤ 6,
y ≥ 0, z ≥ 0.
a) Write out a complete set of first order conditions for this problem.
Answer: The Lagrangian is L = xyz + z − λ0 (x2 + y 2 + z − 6) + λ1 x + λ2 y + λ3 z. This yields three
first-order conditions:
∂L
= yz − 2λ0 x + λ1
∂x
∂L
= xz − 2λ0 y + λ2
0=
∂y
∂L
0=
= xy + 1 − λ0 + λ3
∂z
0=
These can be rewritten as:
yz + λ1 = 2λ0 x
(1)
xz + λ2 = 2λ0 y
(2)
xy + 1 + λ3 = λ0
(3)
b) Determine whether or not the constraint x2 + y 2 + z ≤ 6 is binding at any solution.
Answer: If the constraint does not bind, λ0 = 0 by complementary slackness. In that case, equation
(3) becomes xy + 1 + λ3 = 0, so xy ≤ −1 because λ3 ≥ 0. But the constraints x ≥ 0 and y ≥ 0
imply xy ≥ 0, contradicting xy ≤ −1. We conclude that the constraint must bind at any solution.
c) Find a solution of the first order conditions that includes x = 0.
Answer: If x = 0, (1) becomes yz + λ1 = 0. Since λ1 ≥ 0, either y = 0 or z = 0.
HOMEWORK ASSIGNMENT #6
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√
If z = 0, y = 6 by the constraint from part (a). Then λ2 = 0 by complementary slackness,
√
and (2) becomes 2λ0 y = 0. Since y = 6, this means λ0 = 0. But this cannot be a solution as it
contradicts (3).
If y = 0, z = 6 by part (a). Then λ1 = 0 by (1), λ2 = 0 by (2), λ3 = 0 by complementary slackness,
and λ0 = 1 by (3). In order words, (0, 0, 6) is a possible maximum. Note that f (0, 0, 6) = 6.
d) Find three equations in the three unknowns x, y, z that must be satisfied if x 6= 0 at the solution.
Answer: If x 6= 0, λ1 = 0 by complementary slackness. By (3), λ0 > 0. Multiplying (1) by x and
(2) by z and using complementary slackness, we find 2λ0 x2 = xyz = 2λ0 y 2 . This implies x = y since
λ0 > 0 and x, y ≥ 0. It then implies λ2 = 0 and λ0 = z/2.
We multiply (3) by z, use complementary slackness and λ0 = z/2 to find xyz + z = z 2 /2. Our
three equations are:
6 = x2 + y 2 + z
x=y
2
z /2 = xyz + z
e) Show that x = 1, y = 1, z = 4 satisfies these equations.
Answer: Substitution shows (x, y, z) = (1, 1, 4) solves the equations in part (c). Here f (1, 1, 4) = 6,
the same as in (b).
√ √
Note that even though ( 3, 3, 0) also solves the equations from part (c), it does not solve the
original system. It implies λ0 = 0, contradicting (3).