Question

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

Short answer

The two numbers are \( - 50\) and 50.

Step-by-step solution

Given Data:

1)The difference of the two numbers is 100.

2)The Product is a minimum

Determination of the two numbers.

It is given that the difference between two numbers is 100.

Assumex is smaller and y is larger.

\(\begin{array}{c}y - x = 100\\y = 100 + x\end{array}\)

The function which represents the product of two numbers is:

\(f\left( {x,y} \right) = xy\)……………(1)

Substitute \(y = 100 + x\)in equation (1)

\(\begin{array}{c}f\left( x \right) = x\left( {100 + x} \right)\\ = 100x + {x^2}\end{array}\)

Now, find first derivative of the function \(f\left( x \right) = 100x + {x^2}\)

\(\begin{array}{c}f'\left( x \right) = \left( {100x + {x^2}} \right)'\\ = \left( {100x} \right)' + \left( {{x^2}} \right)'\\ = 100 + 2x\end{array}\)

Solve the equation \(f'\left( x \right) = 0\) to find the critical points.

\(\begin{array}{c}f'\left( x \right) = 100 + 2x\\0 = 100 + 2x\\2x = - 100\\x = - \frac{{100}}{2}\\ = - 50\end{array}\)

Now, find the second derivative of the function \(f\left( x \right)\).

\(\begin{array}{c}f''\left( x \right) = \left( {100 + 2x} \right)'\\ = 100' + 2x'\\ = 0 + 2\end{array}\)

\(f''\left( x \right) = 2 > 0\)

Since, \(f''\left( x \right) > 0,\) there is a minimum.

Now, \(x = - 50\)

Therefore,

\(\begin{array}{l}y = 100 + x\\y = 100 - 50\\y = 50\end{array}\)

Thus, the two numbers are \( - 50\) and 50 whose difference is 100 and whose product is a minimum.