# C 1 a c r 1 x a c 1 12 x dy 2 2a x 1 dy dx 25

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Unformatted text preview: = 6. Maximum 6 at (4/3, 2/3, −4/3), minimum −6 at (−4/3, −2/3, 4/3). 8. 3 = 4xλ, 6 = 8yλ, 2 = 2zλ; 3/(4x) = 3/(4y ) = 1/z thus y = x, z = 4x/3, so 2x2 + 4(x)2 + (4x/3)2 = 70, x2 = 9, x = ±3. Test (−3, −3, −4) and (3,3,4). f (−3, −3, −4) = −35, f (3, 3, 4) = 35. Maximum 35 at (3, 3, 4), minimum −35 at (−3, −3, −4). 9. yz = 2xλ, xz = 2yλ, xy = 2zλ; yz/(2x) = xz/(2y ) = xy/(2z ) thus y 2 = x2 , z 2 = x2 so √ √ √ with x = ±1/ √ y = √ 1/ 3,√ 3, ± and x2 + x2 + x2 = 1, x = ±1/ 3. Test the eight possibilities √ √ √ √ √ z = ±1/ 3 to ﬁnd the maximum is 1/ 3 3 at 1/ 3, 1/ 3, 1/ 3 , 1/ 3, −1/ 3, −1/ 3 , √ √ √ √ √ √ √ −1/ 3, 1/ 3, −1/ 3 , and −1/ 3, −1/ 3, 1/ 3 ; the minimum is −1/ 3 3 at √ √ √ √ √ √ √ √ √ √ √ √ 1/ 3, 1/ 3, −1/ 3 , 1/ 3, −1/ 3, 1/ 3 , −1/ 3, 1/ 3, 1/ 3 , and −1/ 3, −1/ 3, −1/ 3 . 10. 4x3 = λ, 4y 3 = λ, 4z 3 = λ, so x = y = z ; x + y + z = 3x = 1, x = y = z = 1/3, minimum value 1/27, no maximum 11. f (x, y ) = x2 + y 2 ; 2x = 2λ, 2y = −4λ; y = −2x so 2x − 4(−2x) = 3, x = 3/10. The point is (3/10, −3/5). 12. f (x, y ) = (x − 4)2 + (y − 2)2 , g (x, y ) = y − 2x − 3; 2(x − 4) = −2λ, 2(y − 2) = λ; x − 4 = −2(y − 2), x = −2y + 8 so y = 2(−2y + 8) + 3, y = 19/5. The point is (2/5, 19/5). 13. f (x, y, z ) = x2 + y 2 + z 2 ; 2x = λ, 2y = 2λ, 2z = λ; y = 2x, z = x so x + 2(2x) + x = 1, x = 1/6. The point is (1/6, 1/3, 1/6). 14. f (x, y, z ) = (x − 1)2 + (y + 1)2 + (z − 1)2 ; 2(x − 1) = 4λ, 2(y + 1) = 3λ, 2(z − 1) = λ; x = 4z − 3, y = 3z − 4 so 4(4z − 3) + 3(3z − 4) + z = 2, z = 1. The point is (1, −1, 1). 567 Chapter 15 15. f (x, y ) = (x − 1)2 + (y − 2)2 ; 2(x − 1) = 2xλ, 2(y − 2) = 2yλ; (x − 1)/x = (y − 2)/y , y = 2x so x2 + (2x)2 = 45, x = ±3. f (−3, −6) = 80 and f (3, 6) = 20 so (3,6) is closest and (−3, −6) is farthest. 16. f (x, y, z ) = x2 + y 2 + z 2 ; 2x = yλ, 2y = xλ, 2z = −2zλ. If z = 0 then λ = −1 so 2x = −y and 2y = −x, x = y = 0; substitute into xy − z 2 = 1 to get z 2 = −1 which has no real solution. If z = 0 then xy − (0)2 = 1, y = 1/x, and also (from 2x = yλ and 2y = xλ), 2x/y = 2y/x, y 2 = x2 so (1/x)2 = x2 , x4 = 1, x = ±1. Test (1,1,0) and (−1, −1, 0) to see that they are both closest to the origin. 17. f (x, y, z ) = x + y + z , x2 + y 2 + z 2 = 25 where x, y , and z are the components of the vector; 1 = 2xλ√1 = 2yλ, √ = 2zλ; 1/(2x√ = 1/(2y√ = 1/(2z ); y = x,√ = x so x2 + x2 + x2 = 25, , 1 ) ) z √ √ √ √ x = ±5/ 3. f −5/ 3, −5/ 3, −5/ 3 = −5 3 and f 5/ 3, 5/ 3, 5/ 3 = 5 3 so the vector √ is 5(i + j + k)/ 3. 18. x2 + y 2 = 25 is the constraint; 8x − 4y = 2xλ, −4x + 2y = 2yλ; (4x − 2y )/x = (−2x + y )/y , 2x2 + √ − 2y 2 = 0, (2x − y )(x + 2y ) = 0, y = 2x or x = −2√. If y √ 2x then x2 + (2x)2 = 25, 3xy y = √ √√ 5, 2 5 = 0 and x = ± 5. If x = −2y then −2y 2 + y 2 = 25, y = ± 5. T − 5, −2 5 = T √ √ √√ T 2 5, − 5 = T −2 5, 5 = 125. The highest temperature is 125 and the lowest is 0. 19. (a) (b) minimum value −5, maximum value 101/4 15 31.5 -31.5 -27 (c) Find the minimum and maximum values of f (x, y ) = x2 − y subject to the constraint g (x, y ) = x2 + y 2 − 25 = 0, f = λ g , 2xi − j = 2λxi + 2λy j, so solve 2x = 2λx, −1 = 2λy . If x = 0 then y = ±5, f = 5, and if x = 0 then λ = 1, y = −1/2, x2 = 25 − 1/4 = 99/4, √ f = 99/4 + 1/2 = 101/4, so the maximum value of f is 101/4 at (±3 11/2, −1/2) and the minimum value of f is −5 at (0, 5). 20. (a) (b) f ≈ 15 6 5 4 3 2 1 0 0 1 2 3 4 5 6 (d) Set f (x, y ) = x3 + y 3 − 3xy, g (x, y ) = (x − 4)2 + (y − 4)2 − 25; minimize f subject to the constraint g = 0 : f = λg, (3x2 − 3y )i + (3y 2 − 3x)j = 2λ(x − 4)i + 2λ(y − 4)j, so solve 2 (use a CAS) 3x − 3y = 2λ(x − 4), 3y 2 − 3x = 2λ(y − 4); minimum value f = 14.52 at (2.5858, 2.5858) 21. Minimize f = x2 + y 2 + z 2 subject to g (x, y, z ) = x + y + z − 27 = 0. f = λ g , 2xi + 2y j + 2z k = λi + λj + λk, solution x = y = z = 9, minimum value 243 Exercise Set 15.9 568 22. Maximize f (x, y, z ) = xy 2 z 2 subject to g (x, y, z ) = x + y + z − 5 = 0, f = λ g = λ(i + j + k), λ = y 2 z 2 = 2xyz 2 = 2xy 2 z, λ = 0 is impossible, hence x, y, z = 0, and z = y = 2x, 5x − 5 = 0, x = 1, y = z = 2, maximum value 8 at (1, 2, 2) x 23. Minimize f = x2 + y 2 + z 2 subject to√ 2 − yz = 5, f = λ g , 2x = 2xλ, 2y = −zλ, 2z = −yλ. If λ = ±2, √ then y = z = 0, x = ± 5, f = 5; if λ = ±2 then x = 0, and since −yz = 5, √ y = −z = ± 5, f = 10, thus the minimum value is 5 at (± 5, 0, 0). 24. The diagonal of the box must equal the diameter of the sphere so maximize V = xyz or, for convenience, maximize f = V 2 = x2 y 2 z 2 subject to g (x, y, z ) = x2 + y 2 + z 2 − 4a2 = 0, f = λ g , 2xy 2 z 2 = 2λx, 2x2 yz 2 = 2λy, 2x2 y 2√= 2λz . Since V = 0 it follows that x, y, z = 0, hence z √ 2 2 x = ±y = ±z, 3x = 4a , x = ±2a/ 3,...
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