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.

**Expert Answer**

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*