Quiz - Constrained Optimisation

Quiz: Constrained Optimisation July 15, 2014 Substitution Question 1: For each of the following following functions, find the optimum (i.e. maximum or minimum) value of z subject to the given constraint by using direct substitution. 1 2 (a) z = x 3 y 3 subject to the constraint y = 150 − 5x 1 2 Answer: Substituting the constraint into the objective function gives z = x 3 (150 − 5x) 3 . As this is a function of x alone, we can set its first derivative equal to zero to find its maximum, minimum or point of inflection. Thus, −2 −1 −2 2 1 2 dz using the product and chain rules: dx = x 3 (150 − 5x) 3 + x 3 23 (150 − 5x) 3 (−5) = 0. Then, (150 − 5x) 3 13 ∗ x 3 = −1 1 2 1 1 10 3 . Multiplying both sides by x 3 and (150 − 5x) 3 yields (150 − 5x) x 3 10 3 (150 − 5x) 3 = x 3 . This gives x = 10, y = 150 − 5x = 100, and finally z = 46.42. (b) z = x2 + 0.5xy + y 2 subject to the constraint y = 90 − 2x Answer: Substituting the constraint into the objective function gives z = x2 + 0.5x(90 − 2x) + (90 − 2x)2 . This dz is a function of x alone, and dx = 2x + 45 − 2x + 2(90 − 2x)(−2) = 0. From this, x = 39.375, y = 11.25. (c) z = x2 + y 2 − 5xy + 4x + 2y subject to the constraint y = −2x + 100. Answer: As in parts (a) and (b), substitute the constraint directly into the objective function: z = x2 + (−2x + dz 100)2 − 5x(−2x + 100) + 4x + 2(−2x + 100). Thus, z = 15x2 − 900x + (100)2 + 200. Then, dx = 30x − 900 = 0 and x = 30. From the constraint, when x = 30, y = −60 + 100 = 40. Then, x = −3300. Question 2 [More advanved]: Solve each of the problems in question 1 using the Lagrange multiplier method. Answer: 1 2 (a) The Lagrangean expression is L = x 3 y 3 + λ(150 − 5x − y). Calculate the first order conditions (FOC) and set them equal to zero: ∂L ∂x = 31 x −2 3 y 3 − 5λ = 0 (1), which implies 1 3 −1 3 2 2 1 −2 3 y3 15 x 1 3 − λ = 0 (2), which implies 23 x y −1 3 = λ (1a) ∂L ∂x = 32 x y ∂L ∂λ = 150 − 5x − y = 0 (3) Notice that this is the constraint Combining (1a) and (2a), 2 1 −2 3 y3 15 x 1 = 32 x 3 y −1 −1 3 = λ (2a) (4) 1 After multiplying both sides by 15x 3 y 3 , (4) becomes x−1 y = 10 from which y = 10x. Substituting this into (3) gives 150 − 5x − 10x = 0, so x = 10, y = 100, z = 46.416 as we found in question 1. 1 (b) The Lagrangean expression is L = x2 + 0.5xy + y 2 + λ(90 − 2x − y). Use FOCs and set them equal to zero: ∂L ∂x = 2x − 5y + 4 − 2λ = 0 (1), which implies x − 25 y + 2 = λ (1a) ∂L ∂y = 2y − 5x + 2 − λ = 0 (2), which implies 2y − 5x + 2 = λ (2a) ∂L ∂λ = 100 − 2x − y = 0 (3) Combining (1a) and (2a) yields x + 0.25y = 2y + 0.5x, which rearranges to give y = 72 x. Substituting this into (3) gives 16 7 x = 90, so x = 39.375, y = 11.25, z = 1898.438 as before. (c) The Lagrangean expression is L = x2 − 5xy + y 2 + 4x + 2y + λ(100 − 2x − y). Once again, use FOCs and set them equal to zero. ∂L ∂x = 2x − 5y + 4 − 2λ = 0 (1), from which x − ∂L ∂y = 2y − 5x + 2 − λ = 0 (2), from which 2y − 5x + 2 = λ (2a) ∂L ∂λ = 100 − 2x − y = 0 (3) 5 2 = 2y − 5x (1a) Combine (1a) and (2a) as x − 25 y = 2y − 5x, and rearrange to obtain x = 34 y. Substituting this back into (3) gives 100 − 23 y − y = 0. Then, y = 40, x = 43 y = 30, and z = −3300 as before. Question 3: Maximize the following functions, subject to their respective restrictions, using the Lagrange multiplier method. (a) z = 4x2 − 5xy + 6y subject to the constraint x + y = 30 (b) z = −7x2 + 6xy − 9y 2 subject to the constraint 2x + y = 165 (c) z = −3x2 + 40x + 8xy + 288y − 10y 2 subject to the constraint x + 2y = 58 (d) z = 8x2 − 70x − 4xy − 50y + 5y 2 subject to the constraint x + y = 35 Answer: a) The Langrangean is L = 4x2 − 5xy + 6y 2 + λ(x + y − 30). Take the derivatives and set them equal to zero. ∂L = 8x − 5y + λ = 0 ∂x ∂L = 12y − 5x + λ = 0 ∂y ∂L = x + y − 30 = 0 ∂λ Combining the first two and rearranging the third yields: 8x − 5y = 12y − 5x y = 30 − x Solving the system yields the solutions x = 17 and y = 13. 2 b) The Langrangean is L = −7x2 + 6xy − 9y 2 + λ(2x + y − 165). Take the derivatives and set them equal to zero. ∂L = −14x + 6y + 2λ = 0 ⇔ λ = 7x − 3y ∂x ∂L = −18y + 6x + λ = 0 ⇔ λ = −6x + 18y ∂y ∂L = 2x + y − 165 = 0 ∂λ Combining the first two and rearranging the third yields: 7x − 3y = −6x + 18y y = 165 − 2x Solving the system yields the solutions x = 63 and y = 39. c) The Langrangean is L = −3x2 + 40x + 8xy + 288y − 10y 2 + λ(x + 2y − 58). Take the derivatives and set them equal to zero. ∂L = −6x + 40 + 8y + λ = 0 ⇔ λ = 6x − 40 − 8y ∂x ∂L = 8x + 288 − 20y + 2λ = 0 ⇔ λ = 10y − 144 − 4x ∂y ∂L = 2x + y − 165 = 0 ∂λ Combining the first two and rearranging the third yields: 6x − 40 − 8y = 10y − 144 − 4x x = 58 − 2y Solving the system yields the solutions x = 22 and y = 18. d) The Langrangean is L = 8x2 − 70x − 4xy − 50y + 5y 2 + λ(x + y − 35). Take the derivatives and set them equal to zero. ∂L = 16x − 70 − 4y + λ = 0 ∂x ∂L = −4 − 50 + 10yλ = 0 ∂y ∂L = x + y − 35 = 0 ∂λ Combining the first two and rearranging the third yields: 16x − 70 − 4y = −4x − 50 + 10y x = 35 − y Solving the system yields the solutions x = 15 and y = 20. 3