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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 Page 2 (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 Page 3 √ 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).