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Unformatted text preview: maximum volume 8a3 /(3 3). 25. Let x, y , and z be, respectively, the length, width, and height of the box. Minimize f (x, y, z ) = 10(2xy ) + 5(2xz + 2yz ) = 10(2xy + xz + yz ) subject to g (x, y, z ) = xyz − 16 = 0, f = λ g , 20y + 10z = λyz, 20x + 10z = λxz, 10x + 10y = λz . Since V = xyz = 16, x, y, z = 0, thus λz = 20 + 10(z/y ) = 20 + 10(z/x) = 10x + 10y , so x = y . Then z = 16/x2 , thus 20 + 10(16/x3 ) = 20x, x3 + 8 = x4 , the only real solution of which is x = 2,thus x = y = 2, z = 4, minimum value 240. 26. Minimize f (p, q, r) = 2pq +2pr +2qr, subject to g (p, q, r) = p + q + r − 1 = 0, (p,q,r) f = λ 2(q + r) = λ, 2(p + r) = λ, 2(p + q ) = λ, solution p = q = r = 1/3, minimum value 2/3. (p,q,r ) g, 27. Maximize A(a, b, α) = ab sin α subject to g (a, b, α) = 2a + 2b − = 0, (a,b,α) f = λ (a,b,α) g, b sin α = 2λ, a sin α = 2λ, ab cos α = 0 with solution a = b, α = π/2 maximum value if parallelogram is a square. 28. Minimize f (x, y, z ) = xy + 2xz + 2yz subject to g (x, y, z ) = xyz − V = 0, f = λ g , y + 2z = λyz, x + 2z = λxz, 2x + 2y = λxy ; λ = 0 leads to x = y = z = 0, impossible, so solve for λ = 1/z + 2/x = 1/z + 2/y = 2/y + 2/x, so x = y = 2z, x3 = 2V , minimum value 3(2V )2/3 29. (a) Maximize f (α, β, γ ) = cos α cos β cos γ subject to g (α, β, γ ) = α + β + γ − π = 0, f = λ g , − sin α cos β cos γ = λ, −√ α sin β cos γ = λ, − cos α cos β sin γ = λ with solution cos α = β = γ = π/3, maximum value 3 3/8 (b) for example, f (α, β ) = cos α cos β cos(π − α − β ) z x y 30. Find maxima and minima z = x2 + 4y 2 subject to the constraint g (x, y ) = x2 + y 2 − 1 = 0, z = λ g , 2xi + 8y j = 2λxi + 2λy j, solve 2x = 2λx, 8y = 2λy . If y = 0 then λ = 4, x = 0, y 2 = 1 and z = x2 + 4y 2 = 4. If y = 0 then x2 = 1 and z = 1, so the maximum height is obtained for (x, y ) = (0, ±1), z = 4 and the minimum height is z = 1 at (±1, 0). 569 Chapter 15 CHAPTER 15 SUPPLEMENTARY EXERCISES 1. (a) They approximate the proﬁt per unit of any additional sales of the standard or high-resolution monitors, respectively. (b) The rates of change with respect to the two directions x and y , and with respect to time. 3. z = x2 + y 2 = c implies x2 + y 2 = c2 , which is the equation of a circle; x2 + y 2 = c is also the equation of a circle (for c > 0). y y 3 3 x x -3 -3 3 3 -3 -3 z = x2 + y2 5. (b) z= √x2 + y2 f (x, y, z ) = z − x2 − y 2 7. (a) f (ln y, ex ) = eln y ln ex = xy (b) er+s ln rs 8. (a) (b) y 1 y= x y x x –1 1 √ 1 y, wxy = 8xy sec2 (x2 + y 2 ) tan(x2 + y 2 ) + y −1/2 , 2 1 −1/2 1 22 2 22 2 2 , wyx = 8xy sec (x + y ) tan(x + y 2 ) + y −1/2 wy = 2y sec (x + y ) + xy 2 2 9. wx = 2x sec2 (x2 + y 2 ) + 10. ∂w/∂x = 1 1 − sin(x + y ), ∂ 2 w/∂x2 = − − cos(x + y ), x−y (x − y )2 ∂w/∂y = − 1 1 − sin(x + y ), ∂ 2 w/∂y 2 = − − cos(x + y ) = ∂ 2 w/∂x2 x−y (x − y )2 11. Fx = −6xz, Fxx = −6z, Fy = −6yz, Fyy = −6z, Fz = 6z 2 − 3x2 − 3y 2 , Fzz = 12z, Fxx + Fyy + Fzz = −6z − 6z + 12z = 0 12. fx = yz + 2x, fxy = z, fxyz = 1, fxyzx = 0; fz = xy − (1/z ), fzx = y, fzxx = 0, fzxxy = 0 Chapter 15 Supplementary Exercises 13. (a) P = 10 dT 10T dP , = = 12 K/(m2 min) V dT V dt 14. (a) z = 1 − y 2 , slope = ∂z = −2y = 4 ∂y 570 (b) P dV dP = −10 2 = 240 K/(m2 min) dt V dt (b) z = 1 − 4x2 , ∂z = −8x = −8 ∂x 15. x4 − x + y − x3 y = (x3 − 1)(x − y ), limit = −1, continuous 16. x4 − y 4 = x2 − y 2 , limit = lim (x2 − y 2 ) = 0, continuous x2 + y 2 (x,y )→(0,0) √ 1 1 1 2 1 2 √ , √ , v = 0, −1 , w = − √ , − √ = − √ u + √ v, so that 17. Use the unit vectors u = 2 2 5 5 5 5 √ √ 1 1 7 2 2√ Dw f = − √ Duf + √ Dvf = − √ 2 2 + √ (−3) = − √ 5 5 5 5 5 18. (a) n = zx i + zy j − k = 8i + 8j − k, tangent plane 8x + 8y − z = 4 + 8 ln 2, normal line x(t) = 1 + 8t, y (t) = ln 2 + 8t, z (t) = 4 − t (b) n = 3i + 10j − 14k, tangent plane 3x + 10y − 14z = 30, normal line x(t) = 2 + 3t, y (t) = 1 + 10t, z (t) = −1 − 14t 19. The origin is not such a point, so assume that the normal line at (x0 , y0 , z0 ) = (0, 0, 0) passes through the origin, then n = zx i + zy j − k = −y0 i − x0 j − k; the line passes through the origin and is normal to the surface if it has the form r(t) = −y0 ti − x0 tj − tk and (x0 , y0 , z0 ) = (x0 , y0 , 2 − x0 y0 ) lies on the line if −y0 t = x0 , −x0 t = y0 , −t = 2 − x0 y0 , with solutions x0 = y0 = −1, x0 = y0 = 1, x0 = y0 = 0; thus the points are (0, 0, 2), (1, 1, 1), (−1, −1, 1). 2 −1/3 2 −1/3 2 −1/3 −1/3 −1/3 −1/3 2/3 2/3 2/3 x i + y0 j + z0 k, tangent plane x0 x + y0 y + z0 z = x0 + y0 + z0 = 1; 30 3 3 1/3 1/3 1/3 2/3 2/3 2/3 intercepts are x = x0 , y = y0 , z = z0 , sum of squares of intercepts is x0 + y0 + z0 = 1. 20. n = 21. A tangent to the line is 6i + 4j + k, a normal to the surface is n = 18xi + 8y j − k, so solve 18x = 6k, 8y = 4k, −1 = k ; k = −1, x = −1/3, y = −1/2, z = 2 22. ∆w = (1.1)2 (−0.1) − 2(1.1)(−0.1) + (−0.1)2 (1.1) − 0...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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