# Find two numbers whose Difference is $100$ and whose Product is a Minimum

The target of this question is to find two numbers whose sum gives a value of $100$, and the product of those two numbers gives a minimum value. In this question, we will use both algebraic functions and derivatives to find the required two numbers.

Function $f(x,y)$ in mathematics is an expression that describes the relation between two variables $x$ and $y$. In this question, we will assume these two variables:

$x= small value$

$y= large value$

## Numerical Solution

We will now make an equation according to the given data. This equation will be given in the form of “two numbers whose difference is $100$”:

$y – x = 100$

Rearranging the equation gives us:

$y = 100 + x …….. eq.1$

The next equation will show the part of “two numbers whose product is a minimum.” We will use function $f(x,y)$ that will give us the product of x and y:

$f(x,y) = XY……… eq.2$

Substitution of $eq$.$1$ in $eq$.$2$ will give us another expression:

$f(x) = x(100 + x)$

$f(x) = 100x + x^2$

The derivative of a function is the instantaneous rate of change of a function represented by $f'(x)$. We will find the derivatives of the above expression:

$f’ (x) = (100x + x^2)’$

$f’ (x) = 100 + 2x$

Put $f’ (x)$ = $0$ to find the critical points:

$0 = 100 + 2x$

$x = \frac{-100}{2}$

$x = -50$

To check whether $x$=$-50$ is the critical number, we will find the second derivative:

$f’ (x) = 100 + 2x$

$f” (x) = (100 + 2x)’$

$f” (x) = 0 + 2$

$f” (x) = 2 > 0$

A positive value determines that there is a minimum.

Substitution of critical values $x$=$-50$ into the first equation gives us:

$y = 100 + x$

$y = 100 – 50$

$y = 50$

Hence, the solution is $x$=$-50$ and $y$=$50$.

## Example

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

We will assume the two variables as $x$ and $y$:

The product of these two variables will be:

$xy = 100$

$y = \frac{100}{x}$

The sum will be written as:

$sum = x + y$

$sum = x + \frac{100}{x}$

The function will be written as:

$f (x) = x + \frac{100}{x}$

The first derivative of this function gives us:

$f'(x) = 1 – \frac{100}{x^2}$

The second derivative is:

$f” (x) = \frac{200}{x^3}$

Put $f’ (x)$ = $0$ to find the critical points:

$0 = 1 – \frac{100}{x^2}$

$1 =\frac{100}{x^2}$

$x^2 = 100$

$x_1 = 10 , x_2 = -10$

$x_1$=$10$ is a minimum point when $f” (x)$ = $+ve$

$x_2$=$-10$ is the maximum point when $f” (x)$=$-ve$

The sum is minimum at $x$=$10$.

Hence,

$y = \frac{100}{x}$

$y = \frac{100}{10}$

$y = 10$

The two required numbers are $x$=$10$ and $y$=$10$.

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