# Find two positive numbers that satisfy the given requirements. The sum of the first number squared and the second number is 60 and the product is a maximum?

Apr 12, 2017

The numbers are $40$ and $2 \sqrt{5}$. I know these aren't integers (and $\sqrt{5}$ isn't a rational number), but this is the most logical solution to this problem.

#### Explanation:

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

${x}^{2} + y = 60 \to y = 60 - {x}^{2}$

The product will be $P = x y$. Substituting from the first equation, we get:

$P = \left(60 - {x}^{2}\right) x$

$P = - {x}^{3} + 60 x$

We now find the derivative with respect to $x$.

$P ' = - 3 {x}^{2} + 60$

Now determine the critical numbers, which will occur when $P ' = 0$.

$0 = - 3 {x}^{2} + 60$

$0 = - 3 \left({x}^{2} - 20\right)$

$x = \pm \sqrt{20}$

$x = \pm 2 \sqrt{5}$

We must check to make sure $x = + 2 \sqrt{5}$ is indeed a maximum.

Test point $1$: $x = 4$

$P ' \left(4\right) = - 3 {\left(4\right)}^{2} + 60 = \text{positive}$

Test point $2$:$x = 5$

$P ' \left(5\right) = - 3 {\left(5\right)}^{2} + 60 = \text{negative}$

By increasing/decreasing rules, we can conclude that $2 \sqrt{5}$ is a local maximum (this function has no absolute maximum).

This means that $y = 60 - {\left(2 \sqrt{5}\right)}^{2} = 40$.

Hopefully this helps!