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Homework4Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...

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Unformatted text preview: SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #4 Solutions Peter Malkin Section 7.6 Question 2 Let F ( x,y,λ ) = xy- λ (2 x + y- 4). F x = y- 2 λ = 0 ⇔ λ = 1 2 y (1) F y = x- λ = 0 ⇔ λ = x (2) F λ =- (2 x + y- 4) = 0 (3) Using equation (1) and (2) gives 1 2 y = x ⇒ y = 2 x . Putting this into equation (3) gives 2 x + 2 x- 4 = 0 ⇒ x = 4 , so y = 2. Thus, f ( x,y ) = xy is maximum at the point (1 , 2), and the maximum value is f (1 , 2) = 2. Question 6 Let F ( x,y,λ ) = x 2- y 2- λ ( x- 2 y + 6). F x = 2 x- λ = 0 (4) F y =- 2 y + 2 λ = 0 (5) F λ =- ( x- 2 y + 6) = 0 (6) Equation (1) implies λ = 2 x , and substituting this into (2) gives,- 2 y + 4 x = 0, which implies y = 2 x . Substituting this into (3) gives- ( x- 4 x + 6) = 0 ⇒ x = 2 . So, y = 2 x = 4. Thus, f ( x,y ) is maximum at (2 , 4), and the maximum value is f (2 , 4) =- 12. Question 16 Let F ( x,y,λ ) = x 2- 8 x + y 2- 12 y + 48- λ ( x + y- 8). F x = 2 x- 8- λ = 0 (1) F y = 2 y- 12- λ = 0 (2) F λ =- ( x + y- 8) = 0 (3) (1)- (2) : 2 x- 2 y + 4 = 0 ⇒ x- y + 2 = 0 (4) (4) + (3) :- 2 y + 10 = 0 ⇒ y = 5 (5) Substituting y = 5 into (3) gives- ( x + 5- 8) = 0 ⇒ x = 3. So, f ( x,y ) is maximum at (3 , 5), and the maximum value is f (3 , 5) = 3 2- 8 · 3 + 5 2- 12 · 5 + 48 =- 2. 1 Question 18 Let F ( x,y,z,λ ) = x 2 y 2 z 2- λ ( x 2- y 2- z 2- 1). F x = 2 xy 2 z 2- λ 2 x = 0 ⇒ 2 x ( λ- y 2 z 2 ) = 0 ⇒ x = 0 or λ = y 2 z 2 (1) F y = 2 yx 2 z 2- λ 2 y = 0 ⇒ 2 y ( λ- x 2 z 2 ) = 0 ⇒ y = 0 or λ = x 2 z 2 (2) F z = 2 zx 2 y 2- λ 2 z = 0 ⇒ 2 z ( λ- x 2 y 2 ) = 0 ⇒ z = 0 or λ = x 2 y 2 (3) F λ =- ( x 2- y 2- z 2- 1) = 0 (4) Note that we can ignore the cases where x = 0, y = 0, or z = 0 since they do not give a maximum.= 0 since they do not give a maximum....
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This note was uploaded on 01/25/2011 for the course MAT 16C taught by Professor Kouba during the Spring '08 term at UC Davis.

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Homework4Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...

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