Question

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?

Answer 1

The numbers are `40` and `2sqrt(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 -> y = 60 - x^2`

The product will be `P = xy`. Substituting from the first equation, we get:

`P = (60 - x^2)x`

`P = -x^3 + 60x`

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

`P' = -3x^2 + 60`

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

`0 = -3x^2 + 60`

`0 = -3(x^2 - 20)`

`x = +- sqrt(20)`

`x= +- 2sqrt(5)`

We must check to make sure `x = + 2sqrt(5)` is indeed a maximum.

Test point `1`: `x = 4`

`P'(4) = -3(4)^2 + 60 = "positive"`

Test point `2`:`x = 5`

`P'(5) = -3(5)^2 + 60 = "negative"`

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

This means that `y = 60 - (2sqrt(5))^2 = 40`.

Hopefully this helps!