Question

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

Short answer

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

Step-by-step solution

Given Data

1)The sum of two positive numbers is 16.

2)The sum of their squares is the smallest.

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.