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Q. 57

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Calculus
Found in: Page 986
Calculus

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Prove that a square maximizes the area of all rectangles with perimeter P.

∇g = 2, 2∇f = b, aSolving, ∇f = λ∇g we get,a=bwhich is the condition of the rectangle to be square.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1. Given Information.

Given a rectangle with perimeter P. Let a and b be the dimensions of the rectangle.

Step 2. Finding the constraint.

The perimeter is the sum of all sides, which is 2a+2b.

Therefore, the constraint function is:

g(a,b) = 2a+2b.

and it's gradient is:

∇g=2, 2.

The function which maximize the area is:

f(a,b) = A = ab.

and it's gradient is:

∇f = b, a.

Step 3. Using Lagrange's multiplier.

By the method of Lagrange's multiplier, ∇f = λ∇g,So, ∇f = λ2, 2 = 2λ, 2λ.

Now whatever be the value of λ, all the components of ∇f must be same.

So,

b=a

Hence, it is proved that the rectangle must be a square in order to have its area maximum.

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