M408m-Luecke_HW8

# M408m-Luecke_HW8 - romasko (qrr58) – Assignment 8 –...

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Unformatted text preview: romasko (qrr58) – Assignment 8 – luecke – (55035) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the values listed below are ex- treme values of f ( x, y ) = 3 e xy on the region described by the inequality 3 x 2 + 5 y 2 ≤ 1 ? 1. extreme value = 3 e − 1 / √ 15 2. extreme value = 3 e − 1 / 2 √ 15 3. extreme value = 3 e − 5 / √ 15 4. extreme value = 3 e 5 / √ 15 5. extreme value = 3 e 1 / 2 √ 15 correct 6. extreme value = 3 e 1 / √ 15 Explanation: 002 10.0 points Determine the minimum value of f ( x, y, z ) = 2 x 2 + y 2 + 2 z 2 + 3 subject to the constraint 2 x + y + z = 4 . 1. min value = 55 7 2. min value = 50 7 3. min value = 57 7 4. min value = 51 7 5. min value = 53 7 correct Explanation: By the method of Lagrange Multipliers the minimum value of f ( x, y, z ) = 2 x 2 + y 2 + 2 z 2 + 3 subject to the constraint ( † ) 2 x + y + z = 4 will occur at a critical point of the Lagrange function F ( x, y, z, λ ) = 2 x 2 + y 2 + 2 z 2 + 3 + λ (2 x + y + z − 4) . This critical point will be the common solu- tion ( x , y , z , λ ) of the equations ∂F ∂x = 4 x + 2 λ = 0 , ∂F ∂y = 2 y + λ = 0 , ∂F ∂z = 4 z + λ = 0 , and ∂F ∂λ = 2 x + y + z − 4 = 0 . Solving, we see that x = − 1 2 λ , y = − 1 2 λ , z = − 1 4 λ , so 7 4 λ + 4 = 0 , i.e. , λ = − 16 7 . Thus the minimum value of f subject to the constraint ( † ) occurs at parenleftBig 8 7 , 8 7 , 4 7 parenrightBig . Consequently, the minimum value of f subject to the constraint ( † ) is given by min value = 53 7 . 003 10.0 points romasko (qrr58) – Assignment 8 – luecke – (55035) 2 Use Lagrange multipliers to find the shortest distance from the point (1 , − 9 , − 5) to the plane 6 x + 4 y − 10 z = 7. 1. 76 √ 38 13 2. √ 38 3. √ 76 13 4. 13 √ 38 76 correct 5. 7 √ 38 Explanation: keywords: 004 10.0 points Finding the minimum value of f ( x, y ) = 2 x + y − 1 subject to the constraint g ( x, y ) = x 2 + 2 y 2 − 2 = 0 is equivalent to finding the height of the lowest point on the curve of intersection of the graphs of f and g shown in y x z Use Lagrange multipliers to determine this minimum value. 1. min value = − 4 correct 2. min value = − 2 3. min value = − 5 4. min value = − 1 5. min value = − 3 Explanation: The extreme values occur at solutions of ( ∇ f )( x, y ) = λ ( ∇ g )( x, y ) . Now ( ∇ f )( x, y ) = ( 2 , 1 ) , while ( ∇ g )( x, y ) = ( 2 x, 4 y ) . Thus 2 = 2 λx , 1 = 4 λy , and so λ = 1 x = 1 4 y , i . e ., x = 4 y ....
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## This note was uploaded on 06/15/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.

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M408m-Luecke_HW8 - romasko (qrr58) – Assignment 8 –...

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