Calculus question

Question

Find 2 positive numbers whose sum is 18 and the product of the first number and the square of the other is a maximum.

 

Help!?

Answer 1

Let x = first number

Let y = other number

 

x + y = 18                       eq1

 

Product = xy²                    eq2

 

 

Substitute eq1 into eq2.

 

Product = x(18 - x)²

            = x(324 - 36x + x²)

            = 324x - 36x² + x³

 

Next, find the derivative of Product and set it equal to zero.

 

324 - 102x + 3x² = 0

 

3(108 - 34x + x²) = 0

 

 

Solve the quadratic equation to solve for x.  You will get two values of x.  If there is only one positive solution, then accept that solution.  If there are two positive solutions, then you need to evaluate the derivative around each solution.

The derivative before the chosen x value should be positive, and the derivative after the chosen x value should be negative. 

 

 

Once you have your x value determined, use it to solve for y.

Answer 2

Let x be the first number and let y be the second number

 

Then x+y=18.  So,y = 18-x.

 

Maximize P = xy² = x(18-x)²

 

P = x(324 - 36x + x²) = x³ - 36x² + 324x

 

P' = 3x² - 72x + 324 = 3(x² - 24x + 108)

 

                               = 3(x-6)(x-18)

 

P' = 0 when x = 6 or x = 18

 

If x = 18, then y = 0.  This is not possible since y must be positive.  

 

So, x = 6

 

If 0<x<6, then P'>0.  So, P is increasing.

 

If 6<x<18, then P'<0.  So, P is decreasing.  

 

Therefore, there is a relative and absolute max when x = 6 and y = 12.