 # 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 $y$=$10$.

Image/Mathematical drawings are created in Geogebra