Solutions to Homework 7.5

14.3

#4 Let g(x, y) = 4x^{2} +y^{2}, then ∇f = λ∇g gives the equations y = λ(8x), x = λ(2y). Eliminating λ yields y^{2} = 4x^{2} and plugging this back into the constraint gives 8x^{2} = 8, so that x = −1, 1 and the corresponding y = −2, 2. Thus our critical points are (-1,-2), (-1,2), (1,-2) and (1,2) and the values of f at these points are 2, -2, -2 and 2 respectively. So our maximum value is 2 and our minimum -2.

#5 Let g(x, y) = x^{4} + y^{4}, then ∇f = λ∇g gives the equations 2x = λ(4x^{3}), y = λ(4y^{3}). Eliminating λ yields x = 0, y = 0 or y^{2} = x^{2} and plugging this back into the constraint gives y^{4} = 2, x^{4} = 2 and 2x^{4} = 2 respectively. Our critical points are thus (0, ± ^{4} \sqrt{2}), (± ^{4} \sqrt{2}), (1, ±1) and (−1, ±1) and computing the function values we get that the first 4 points have f(x, y) = ^{4} \sqrt{2} and the last 4 points have f(x, y) = 2. Hence our constrained max is 2 and our constrained min is ^{4} \sqrt{2}.

#11 a) These look like parts of hyperbolas.

b) s = 4(100 − l).

c) This line should go from s-intercept 400 to l-intercept 100 and touch indifference curve II at (50,200).

d) First note that curve III is the most preferable, and I the least preferable, so we would like to be on the outermost curve that our constarint allows. That curve is II, and we barely touch it; all more preferable curves will therefore miss the constraint and our constrained maximum occurs at (50,200). Hence working 50hrs and taking 50 hrs off per week is the optimum strategy in the given situation.

#16

a) The constraint is c = 10W + 20K = 3000. So ∇f = λ∇c gives the equations \frac{9}{2}W^{−1/4}K^{1/4} = λ10 and \frac{3}{2}W^{3/4}K^{−3/4} = λ20. Solving yields W = 6K and plugging this into the equation for c we get that W = 225 and K = 37.5. Also λ = 0.2875.

b) q_{W} (225, 37.5) = 2.875 and q_{K}(225, 37.5) = 5.750 and the quiotient of these two numbers is .5=10/20=cost of W/cost of K.

c) Our constraint changes to c = 3001, solving yields W′ = 225.075 and K′ = 37.513 and thus q′ = 862.86 which is roughly .29 bigger then the maximum from part a). Note that λ ≈ 0.29, so the increase in q was as we expected.