What size square should be cut out to create a box with greatest volume? What is the maximum volume?

Question

An open box is to be made from a piece of plastic 20 by 20 inches by cutting out squares of equal size from the corners and bending up the sizes. 

Answer 1

This is a classic optimization problem in calculus.    Let s be the length of the sides of the squares to be cut out.

 

Then the dimensions of the open box will be 20 - 2s by 20 -2s by s  .

The volume is thus  V  = (20 - 2s)² s

 

To find the value of s which maximizes this, compute the derivative   dV/ds and set it equal to zero.

 

 

Using the product rule,  power rule and chain rule   dV/ds  = (20 - 2s)² - 4 s (20 - 2s)

 

Setting this equal to zero gives  (20 - 2s) - 4 s = 0   with the solution

 

   s =  10/3

 

The maximum volume is then  (20 - 20/3)² (10/3) =  16000/27

 

 

Answer 2

Length = 20 - 2x

Width = 20 - 2x

Height = x

 

Volume V(x) = (20 - 2x)(20-2x)x

 

                   = (400 - 80x + 4x^2)x

 

                   = (4x^2 - 80x + 400)x

 

                   = 4x^3 - 80x^2 + 400x

 

 

Maximizing:

 0 = dV/dx = V'(x) = 12x^2 - 160x + 400

 

                      3x^2 - 40x + 100 <---- reduced by factor of 4

 

                      (  3x      -  10   )( x   -     10   )

 

                   3x-10 = 0   or x-10 = 0

 

                     x = 10/3  or x = 10

 

  x=10 results in non-positive measures, namely 20 - 2(10) = 0 for length and width

 

  x=10/3 yields the max volume

  Max Volume is (10/3)(20 - 2(10/3)) ( 20 - 2(10/3))

                       (10/3)( 20 - 20/3) ( 20 - 20/3)

                      (10/3)(  (60-20)/3) ( (60-20)/3)

                     (10/3)( (40/3)(40/3)

                     16000/27=592 and 592/666