Question

Find two numbers whose difference is 100 and whose product is a minimum.

Short answer

The two numbers are \[50{\rm{ and }} - 50\].

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:

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

Optimizing the function according to a given condition.

Let one of the numbers be \[x\]. Then another number will be: \[100 + x\].

Here, the product of the two will be:

\[\begin{aligned}{c}P\left( x \right) = x\left( {100 + x} \right)\\ = 100x + {x^2}\end{aligned}\]

For minimization, we have:

\[\begin{aligned}{c}P'\left( x \right) = 100 + 2x\\100 + 2x = 0\\x = - 50\end{aligned}\]

Also, the second derivative, that is:

\[P''\left( x \right) = 2 > 0\]

Therefore, the function will have an absolute minimum value at \[x = - 50\].

So, we have the second number as:

\[\begin{aligned}{c}100 + x = 100 - 50\\ = 50\end{aligned}\]

Hence, the required numbers are \[50{\rm{ and }} - 50\].