Question

Find two positive numbers whose product is 100 and whose sum is a minimum.

Short answer

The two numbers are \({\rm{10 and }}10\).

Step-by-step solution

Optimization of a Function

The Optimization 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 a given condition.

Let one of the numbers be \(x\). Then another number will be: \(\frac{{100}}{x}\).

Here, the product of the two will be:

\(\begin{aligned}{c}S\left( x \right) = x + \frac{{100}}{x}\\ = \frac{{{x^2} + 100}}{x}\end{aligned}\)

For minimization, we have:

\(\begin{array}{c}S'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{{x^2} + 100}}{x}} \right)\\\frac{d}{{dx}}\left( {\frac{{{x^2} + 100}}{x}} \right) = 0\\\frac{{{x^2} - 100}}{{{x^2}}} = 0\\x = \pm 10\\x = 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,............\left\{ {{\rm{only positive value}}} \right\}\end{array}\)

Also, the second derivative, that is:

\(S''\left( x \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,............\left\{ {\forall x \in \left( {0,10} \right)} \right\}\)

Therefore, the function will have an absolute minimum value at \(x = 10\).

So, we have the second number as:

\(\begin{array}{c}\frac{{100}}{x} = \frac{{100}}{{10}}\\ = 10\end{array}\)

Hence, the required numbers are \(10{\rm{ and }}10\).