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Q. 27

Expert-verified

Calculus

Found in: Page 985

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.
$f (x, y) = x y when x^{2} + 4 y^{2} = 16$

The maximum value of the function is $4$ and the minimum value is $- 4$ and both exist as the constraint is a bounded and closed ellipse.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Given function is $f (x, y) = x y .$

Given constraint is $x^{2} + 4 y^{2} = 16 .$

Step 2. critical points of the function.

Gradients of function.

$\nabla f (x, y) = y i + x j \nabla g (x, y) = 2 x i + 8 z j$

Use the method of Lagrange multipliers.

$\nabla f (x, y) = λ \nabla g (x, y) y i + x j = λ (2 x i + 8 y j) y i + x j = 2 λ x i + 8 λ y j$

Compare terms.

$y = 2 λ x \Rightarrow λ = \frac{y}{2 x} x = 8 λ y \Rightarrow λ = \frac{x}{8 y} so x^{2} = 4 y^{2}$

substitute $x^{2} = 4 y^{2}$ in constraint.

role="math" localid="1649893915369" $x^{2} + 4 y^{2} = 16 4 y^{2} + 4 y^{2} = 16 8 y^{2} = 16 y = \pm \sqrt{2} x = \pm 2 \sqrt{2}$

so critical points are role="math" localid="1649893960368" $(- 2 \sqrt{2}, - \sqrt{2}), (2 \sqrt{2}, - \sqrt{2}), (- 2 \sqrt{2}, \sqrt{2}), & (2 \sqrt{2}, \sqrt{2}) .$

Step 3. maximum and minimum of a function.

Find function value at $(- 2 \sqrt{2}, - \sqrt{2}), (2 \sqrt{2}, - \sqrt{2}), (- 2 \sqrt{2}, \sqrt{2}), & (2 \sqrt{2}, \sqrt{2}) .$

$f (x, y) = x y f (- 2 \sqrt{2}, - \sqrt{2}) = (- 2 \sqrt{2}) (- \sqrt{2}) f (- 2 \sqrt{2}, - \sqrt{2}) = 4$

similarly,

$f (2 \sqrt{2}, - \sqrt{2}) = - 4 f (- 2 \sqrt{2}, \sqrt{2}) = - 4 f (- 2 \sqrt{2}, - \sqrt{2}) = 4$

So the maximum value of the function is $4$ and the minimum value is $- 4 .$

As constraint is bounded and closed ellipse so maximum and minimum both exist.

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