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HW12S - Version 100 Homework 12 Gilbert(59825 This...

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Version 100 – Homework 12 – Gilbert – (59825) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. I’ve put a few numerical free response questions on this assignment to test this feature. Return your answer as a deci- mal to, say, 3 places. For the answer to be graded as correct is has to be within 1% of the correct answer. 001 10.0 points Determine the minimum value of f ( x, y, z ) = 2 x 2 + y 2 + 2 z 2 + 6 subject to the constraint 2 x + 2 y + z = 6 . 1. min value = 148 13 2. min value = 147 13 3. min value = 150 13 correct 4. min value = 152 13 5. min value = 154 13 Explanation: By the method of Lagrange Multipliers the minimum value of f ( x, y, z ) = 2 x 2 + y 2 + 2 z 2 + 6 subject to the constraint ( ) 2 x + 2 y + z = 6 will occur at a critical point of the Lagrange function F ( x, y, z, λ ) = 2 x 2 + y 2 + 2 z 2 + 6 + λ (2 x + 2 y + z - 6) . This critical point will be the common solu- tion ( x 0 , y 0 , z 0 , λ 0 ) of the equations ∂F ∂x = 4 x + 2 λ = 0 , ∂F ∂y = 2 y + 2 λ = 0 , ∂F ∂z = 4 z + λ = 0 , and ∂F ∂λ = 2 x + 2 y + z - 6 = 0 . Solving, we see that x 0 = - 1 2 λ 0 , y 0 = - λ 0 , z 0 = - 1 4 λ 0 , so 13 4 λ 0 + 6 = 0 , i.e. , λ 0 = - 24 13 . Thus the minimum value of f subject to the constraint ( ) occurs at parenleftBig 12 13 , 24 13 , 6 13 parenrightBig . Consequently, the minimum value of f subject to the constraint ( ) is given by min value = 150 13 . 002 10.0 points The temperature T at a point ( x, y, z ) on the surface x 2 + y 2 + z 2 = 48 is given by T ( x, y, z ) = x + y + z in degrees centigrade. Find the maximum temperature on this surface. Correct answer: 12. Explanation:
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Version 100 – Homework 12 – Gilbert – (59825) 2 We have to maximize the function T sub- ject to the constraint x 2 + y 2 + z 2 = 48 . The corresponding Lagrange function is F ( x, y, z, λ ) = x + y + z + λ ( x 2 + y 2 + z 2 - 48) . At the critical point of F , therefore, F x = 1 + 2 λx = 0 , F y = 1 + 2 λy = 0 , F z = 1 + 2 λz = 0 , F λ = x 2 + y 2 + z 2 - 48 = 0 . Thus x = y = z = radicalbig 48 / 3 = 4 . Substituting in the original definition of T we see that T max = 12 C . 003 10.0 points Use Lagrange multipliers to find the vol- ume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid 9 x 2 + 3 y 2 + 4 z 2 = 1 . 1. volume = 4 9 2. volume = 1 9 3. volume = 4 27 correct 4. volume = 2 27 5. volume = 2 9 Explanation: The rectangular box will be centered at the origin, so if the corner lying in the first octant is P ( x, y, z ), then the box will have sidelengths 2 x, 2 y and 2 z . Thus the box will
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