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Dieter Balkenborg, Giovanni Caggiano BEE1020 — Basic Mathematical Economics Axel Dreher, Iannis Krassas Department of Economics Homework — Solutions 10-11/02/2004 Week 15 University of Exeter Exercise 1 Find the critical point of the functions a) z b) z c) z = f (x, y) = xy − 2x + 3y − 6 = g (x, y) = 2x2 + 2xy − 6x + 5y 2 − 6y + 5 = h (x, y) = −2x2 − 2xy + 6x − 5y 2 − 5 Determine whether they are troughs, peaks or saddlepoints. Solution 1 a) ∂z ∂x ∂z ∂y = y−2=0⇔y =2 = x + 3 = 0 ⇔ x = −3 The critical point is (x, y) = (3, 2). det H = · 0 1 1 0 ¸ = −1 < 0 The critical point is hence a saddle point. b) ∂z ∂x ∂z ∂y = 4x + 2y − 6 = 0 = 2x + 10y − 6 = 0| × 2 4x + 2y = 6| − 4x + 20y = 12| + −18y = −6 Hence y = 13 , 4x = 6 − 2 3 = 16 3 , x = 43 . The critical point is (x, y) = · ¸ 4 2 det H = = 40 − 4 = 36 > 0 2 10 2 ¢ 1 , 3 3 . Since ¡4 ∂ z the critical point is not a saddle point. Since ∂x 2 = 4 > 0 it is a trough. (In fact, the function is strictly convex and the critical point a global minimum.) c) ∂z ∂x ∂z ∂y = −4x − 2y + 6 = 0 = −2x − 10y = 0| × (−2) −4x − 2y = −6| + 4x + 20y = 0| + 18y = −6 Hence y = − 13 , 4x = 6 + 2 3 x = 53 . The critical point is again (x, y) = · ¸ −4 −2 det H = = 40 − 4 = 36 > 0 −2 −10 = 20 3 , ¡5 1 3, −3 ¢ . Since 2 ∂ z the critical point is not a saddle point. Since ∂x 2 = −4 < 0 it is a peak. (In fact, the function is strictly concave and the critical point a global maximum.) Exercise 2 A consumer has the utility function u (x, y) = 5x2 + 6xy + y 2 + 38x + 18y a) Determine the marginal utility for the two commodities. Is more always better for the consumer? b) The consumer has a budget of $40. A unit of the first commodity costs $10 and a unit of the second $5. Write down the budget equation. c) The consumer wants to maximize his utility subject to his budget constraint. Write down the Lagrangian for this problem. d) Calculate the first order conditions for a critical point of the Lagrangian. e) Derive a linear equation to be satisfied by a critical point that does not involve the Lagrange multiplier λ. f) Use the equation from e) and the budget equation to find the constrained optimum. Solution 2 a) marginal utilities are ∂u ∂x ∂u ∂y = 10x + 6y + 38 = 6x + 2y + 18 These are positive when x and y are positive. b) budget constraint: 10x + 5y ≤ 40 budget equation: 10x + 5y = 40 c) Lagrangian: L (x, y) = u (x, y) − λ [10x + 5y − 40] = 5x2 + 6xy + y 2 + 38x + 18y − λ [10x + 5y − 40] d) conditions for critical point ∂L ∂x ∂L ∂y = 10x + 6y + 38 − 10λ = 0 = 6x + 2y + 18 − 5λ = 0 2 e) Eliminate λ: Divide first equation by 2 and then subtract second one. 5x +3y +19 −5λ = 0 6x +2y +18 −5λ = 0 −x +y +1 = 0 f) solve simultaneous system −x + y + 1 = 0 from e) 10x + 5y = 40 budget equation from b) division of the budget equation by 5 yields −x +y +1 = 0 2x +y = 8 |− −3x +1 = −8 −3x = −9 The candidate for the maximum is x = 3, y = x − 1 = 2. 3