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Q 4E

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Essential Calculus: Early Transcendentals
Found in: Page 238
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

The sum of two positive numbers is 16. What is the smallest possible value of the sum of their squares?

The smallest possible value of the sum of the squares of the two positive numbers is 128.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

1)The sum of two positive numbers is 16.

2)The sum of their squares is the smallest.

Step 2: Determination of the two numbers

Let the numbers be x and y.

It is given that the sum of the two numbers is 16.

Therefore, it can be written as:

\(\begin{aligned}{c}x + y &= 16\\y &= 16 - x\end{aligned}\)

The function which represents the sum of the squares of two numbers is:

\(f\left( {x,y} \right) = {x^2} + {y^2}\) ……………(1)

Substitute \(y = 16 - x\)in equation (1)

\(f\left( x \right) = {x^2} + {\left( {16 - x} \right)^2}\)

\(f\left( x \right) = 2{x^2} - 32x + 256\)

A minimum of \(f\left( x \right)\) is obtained at \(f'\left( x \right) = 0\)

\(f\left( x \right) = 2{x^2} - 32x + 256\)

\(f'\left( x \right) = 4x - 32\)

Substitute \(f'\left( x \right) = 0\) as:

\(\begin{aligned}{c}4x - 32 &= 0\\4x &= 32\\x &= 8\end{aligned}\)

Therefore,

\(\begin{aligned}{c}y &= 16 - x\\y &= 16 - 8\\ &= 8\end{aligned}\)

And the minimum possible value the sum of the squares is:

\(\begin{aligned}{c}{8^2} + {8^2} &= 64 + 64\\ &= 128\end{aligned}\)

Thus, the smallest possible value of the sum of the squares of the two positive numbers is 128.

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