# 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?

##### 1 Answer

The numbers are

#### Explanation:

Let the numbers be

#x^2 + y = 60 -> y = 60 - x^2#

The product will be

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

#P = -x^3 + 60x#

We now find the derivative with respect to

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

Now determine the critical numbers, which will occur when

#0 = -3x^2 + 60#

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

#x = +- sqrt(20)#

#x= +- 2sqrt(5)#

We must check to make sure

**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

This means that

Hopefully this helps!