Homework Assignment #4 13.11 Write the following linear functions in matrix form: a) f (x1 , x2 , x3 ) = 2x1 − 3x2 + 5x3 ; b) f (x1 , x2 ) = (2x1 − 3x2 , x1 − 4x2 , x1 ); c) f (x1 , x2 , x3 ) = (x1 − x3 , 2x1 + 3x2 − 6x3 , x3 + 2x2 ). Answer: x1 2x1 − 3x2 + 5x3 = (2, −3, 5) x2 x3 2 −3 x1 (2x1 − 3x2 , x1 − 4x2 , x1 )T = 1 −4 x2 1 0 x1 1 0 −1 T x2 (x1 − x3 , 2x1 + 3x2 − 6x3 , x3 + 2x2 ) = 2 3 −6 x3 0 2 1 (a) (b) (c) 13.12 Write the following quadratic forms in matrix form: a) x21 − 2x1 x2 + x22 . b) 5x21 − 10x1 x2 − x22 . c) x21 + 2x22 + 3x23 + 4x1 x2 − 6x1 x3 + 8x2 x3 . Answer: If we require the matrices to be symmetric, the solutions are: 1 2 −3 1 −1 5 −5 xT x, xT x, xT 2 2 4 x. −1 1 −5 −1 −3 4 3 13.20 Prove carefully that h : Rk → R1 defined by h(x1 , x2 , . . . , xk ) = xi is continuous on Rk . Conclude that any monomial g(x1 , x2 , . . . , xk ) = cxn1 1 xn2 2 · · · xnk k is continuous on Rk and that any polynomial from Rk to Rm is continuous on Rk . Answer: Suppose xn → x. Then |h(xn ) − h(x)| = |xni − xi | ≤ kx− xk. Since xn → x, for any > 0 we can choose N such that kxn − xk < for all n ≥ N . Then |h(xn ) − h(x)| < for all n ≥ N , showing h(xn ) → h(x), which establishes continuity of h. We also known that multiplying a continuous function by a continuous function (including constant functions) yields a continuous function.It follows that monomial g from Rk to R1 is continuous. Since the addition of continuous functions yields a continuous function, any real-valued polynomial on Rk is continuous. Finally, since each component is continuous, and polynomial from Rk to Rm is continuous. 14.5 The demand function Q1 = K1 P1a11 P2a12 I b1 is called a constant elasticity demand function. a) Compute the three elasticities (own price, cross price, and income) and show that they are all constants. Answer: The own price elasticity is (P1 /Q1 )∂Q1 /∂P1 = (P1 /Q1 )a11 Q1 /P1 = a11 . = (P1 /Q1 )a11 K1 P1a11 −1 P2a12 I b1 HOMEWORK ASSIGNMENT #4 Page 2 The cross price elasticity = (P2 /Q1 )a12 Q1 /P2 = a12 . is (P2 /Q1 )∂Q1 /∂P2 = (P2 /Q1 )a12 K1 P1a11 P2a12 −1 I b1 The income elasticity is (I/Q1 )∂Q1 /∂I = (I/Q1 )b1 K1 P1a11 P2a12 I b1 −1 = (I/Q1 )a12 Q1 /I = b1 . b) What are reasonable ranges for the four parameters in Q1 ? Answer: Demand should be positive, requiring K1 > 0. The own price elasticity should be negative, requiring a11 < 0. Since goods can be either substitutes or complements, we can’t sign a12 , and because good one can be either normal or inferior, we can’t sign b1 . 14.8 Use differentials to approximate each of the following: a) f (x, y) = x4 + 2x2 y 2 + xy 4 + 10y at x = 10.36 and y = 1.04; Answer: df = (4x3 + 4xy 2 + y 4 ) dx + (4x2 y + 4xy 3 + 10) dy. We evaluate this at the point (10, 1), obtaining df = 4041 dx + 450 dy. Replacing dx by ∆x = .36 and dy by ∆y = .04, we find ∆f ≈ 4041(.36) + 450(.04) = 1454.76 + 18 = 1472.76. Since f (10, 1) = 10220, the approximate value of the function is 11692.76 while the actual value is about 11744.3. b) f (x, y) = 6x2/3 y 1/2 at x = 998 and y = 101.5; Answer: Here we approximate around (x, y) = (1000, 100), where f (1000, 100) = 6000. Then df = 4x−1/3 y 1/2 dx + 3x2/3 y −1/2 dy. Evaluating at (1000, 100), we find df = 4 dx + 30 dy. Replacing dx by ∆x = −2 and dy by ∆y = 1.5, we find ∆f ≈ −8 + 45 = 37, so f (998, 101.5) ≈ 6037. In comparison, the actual value is about 6036.77. p c) f (x, y, z) = x1/2 + y 1/3 + 5z 2 at x = 4.2, y = 7.95, and z = 1.02. Answer: Here we evaluate around the point (x, y, z) = (4, 8, 1). Note that f (4, 8, 1) = 3. Then 1 df = p 2 x1/2 + y 1/3 + 5z 2 1 −2/3 1 −1/2 x dx + y dy + 10z dz . 2 3 1 dy + 10 dz]. We use ∆x = 0.2, ∆y = −0.05, and ∆z = 0.02. At (4, 8, 1), this becomes 16 [ 14 dx + 12 1 This yields ∆f ≈ 6 [0.05 − 0.025 + 0.2] = 0.0375, so f (4.2, 7.95, 1.02) ≈ 3.0375, compared to the actual value of about 3.041. 14.16 The university’s solar car is racing on Australia’s roads on the path (x, y) = (et + 5t2 , t4 − 4t). What are the coordinates of the car when its velocity vector is pointing parallel to the x-axis? Answer: The velocity vector is (et +10t, 4t3 −4). This will be parallel to the x-axis whenever 4t3 −4 = 0, i.e., t = 1. At t = 1, the coordinates of the car are (x, y) = (5 + e, −3).

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