Question

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

Short answer

The two numbers are \({\rm{8 and }}8\). The smallest possible value of sum of their squares is \(128\).

Step-by-step solution

Optimization of a Function

TheOptimization of a Function\(f\left( x \right)\)for any real value of\(c\)can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{array}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{array} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{array}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{array} \right.} \right.\end{aligned}\).

Optimizing the function according to given condition.

Let one of the numbers be \(x\). Then other number will be: \(16 - x\).

Here, the sum of squares of the two will be:

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

For minimization, we have:

\(\begin{aligned}{c}S'\left( x \right) = \frac{d}{{dx}}\left( {{x^2} + {{\left( {16 - x} \right)}^2}} \right)\\\frac{d}{{dx}}\left( {{x^2} + {{\left( {16 - x} \right)}^2}} \right) = 0\\2x - 2\left( {16 - x} \right) = 0\\4x - 32 = 0\\x = 8\end{aligned}\)

Also, second derivative, that is:

\(S''\left( x \right) > 0\)

Therefore, the function will have absolute minimum value at \(x = 8\).

So, we have the second number as:

\(\begin{array}{c}16 - x = 16 - 8\\ = 8\end{array}\)

And the smallest possible value of their squares is:

\(\begin{array}{c}S = {8^2} + {8^2}\\ = 128\end{array}\)

Hence, the required number is \(128\).