# What are two positive numbers whose sum of the first number squared and the second number is 54 and the product is a maximum?

May 9, 2018

$3 \sqrt{2} \mathmr{and} 36$

#### Explanation:

Let the numbers be $w$ and $x$.

${x}^{2} + w = 54$

We want to find

$P = w x$

We can rearrange the original equation to be $w = 54 - {x}^{2}$. Substituting we get

$P = \left(54 - {x}^{2}\right) x$

$P = 54 x - {x}^{3}$

Now take the derivative with respect to $x$.

$P ' = 54 - 3 {x}^{2}$

Let $P ' = 0$.

$0 = 54 - 3 {x}^{2}$

$3 {x}^{2} = 54$

$x = \pm \sqrt{18} = \pm 3 \sqrt{2}$

But since we're given that the numbers have to be positive, we can only accept $x = 3 \sqrt{2}$. Now we verify that this is indeed a maximum.

At $x = 3$, the derivative is positive.

At $x = 5$, the derivative is negative.

Therefore, $x = 3 \sqrt{2}$ and $54 - {\left(3 \sqrt{2}\right)}^{2} = 36$ give a maximum product when multiplied.

Hopefully this helps!