Optimization

Question

The sum of a number and the cube of a second number is 32. Find the maximum product of the numbers. 

Answer 1

Hello Cortni,

From the first sentence, we get

x + y³ = 32,

→ x = 32 - y³.

 

From the second sentence, we want to maximize the product of the two numbers.  So, let

P = x * y

→ P = (32-y³)(y), by substituting the 'x' out,

→ P = 32y - y⁴.

 

To maximize this product, we need to take the derivative, set it equal to zero, solve for y, and do line analysis.

→ P' = 32 - 4y³ = 0,

→ 4y³ = 32,

→ y³ = 8

→ y = 2.

 

For the line analysis, test a point on either side of 2.

If y = 0

→ P'(0) = 32 > 0 : a positive value

 

If y = 3

→ P'(3) = -76 < 0 : a negative value

This verifies that at y = 2, we will get a maximum product.

 

So, for y = 2,

→ x = 32 - (2)³ = 32 - 8 = 24

 

→ P_max = (24)(2) = 48.

 

Thank you for the question Cortni, and have a great day!

Michael E.

Answer 2

x + y³ = 32, so x = 32 - y³

 

Maximize xy = (32 - y³)(y) = 32y - y⁴

 

Let f(y) = 32y - y⁴

 

     f'(y) = 32 - 4y³

 

     f'(y) = 0 when y = 2

 

If y < 2, then f'(y) > 0, so f(y) is increasing

 

If y > 2, then f'(y) < 0, so f(y) is decreasing

 

f(y) is maximized when y = 2, and x = 32 - y³ = 24

 

Maximum product = xy = 48