Q. 27 In Exercises, find the maximum a... [FREE SOLUTION]

Chapter 12: Q. 27 (page 985)

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.
$f (x, y) = x y when x^{2} + 4 y^{2} = 16$

Short Answer

Expert verified

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.

Step by step solution

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.

$鈭� f (x, y) = y i + x j 鈭� g (x, y) = 2 x i + 8 z j$

Use the method of Lagrange multipliers.

$鈭� f (x, y) = 位鈭� g (x, y) y i + x j = 位 (2 x i + 8 y j) y i + x j = 2 位 x i + 8 位 y j$

Compare terms.

$y = 2 位 x 鈬� 位 = \frac{y}{2 x} x = 8 位 y 鈬� 位 = \frac{x}{8 y} so x^{2} = 4 y^{2}$

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

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

so critical points arerole="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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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.
$f (x, y) = x y when x^{2} + 4 y^{2} = 16$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. critical points of the function.

Step 3. maximum and minimum of a function.

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

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)=xywhenx2+4y2=16

Short Answer

Step by step solution

Step 1. Given information.

Step 2. critical points of the function.

Step 3. maximum and minimum of a function.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Theoretical and Mathematical Physics

Calculus

Probability and Statistics

Pure Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

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.
$f (x, y) = x y when x^{2} + 4 y^{2} = 16$