test2sol

test2sol - Math 263 Test 2 Solutions 1 Use Lagrange...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 263 Test 2 Solutions 1. Use Lagrange multipliers to find the maximum and minimum values of the function f ( x, y ) = 1 x + 1 y subject to the constraint 1 x 2 + 1 y 2 = 1 . Solution: We define the function g ( x, y ) = 1 x 2 + 1 y 2- 1 and using Lagrange multipliers er get that critical points of f subject to the constraint g = 0 satisfy ∇ f ( x, y ) = λ ∇ g ( x, y ) and g ( x, y ) = 0 . Now ∇ f ( x, y ) = h- 1 x 2 ,- 1 y 2 i and ∇ g ( x, y ) = h- 2 x 3 ,- 2 y 3 i . Hence we have to solve the system of equations- 1 x 2 =- 2 λ x 3 ,- 1 y 2 =- 2 λ y 3 and 1 x 2 + 1 y 2 = 1 . From the first two equations we get x = 2 λ = y. Inserting this into the third equation gives x = ± √ 2 . Hence the maximum value of f is f ( √ 2 , √ 2) = √ 2 and the minimum value is f (- √ 2 ,- √ 2) =- √ 2 . 1 2. Consider the integral Z 1 Z 1 x 2 x 3 e 2 y 3 dy dx. (a) Sketch the region of integration....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

test2sol - Math 263 Test 2 Solutions 1 Use Lagrange...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online