Jane needs open-topped boxes to store her excess inventory at year's end. She purchases large rectangles of thick cardboard with a length of 12 inches and width of 8 inches to make the boxes.
She is interested in maximizing the volume of the boxes and wants to know what size squares to
cut out at each corner of the cardboard (which will allow the corners to be folded up to form the
box) to do this. What is the maximum volume?

x = 1,6 in   ( the side of the corner squares )

V(max) = 67,58 in³

Step-by-step explanation:

The cardboard is:

L = 12 in          w  =  8 in

Let´s call "x" the side of the square from the corner:

Then the sides of the base of the open box are:

( L - 2*x )    and    ( w - 2*x )     and   x is the height

( 12  -  2*x )  and  ( w  - 2*x )

V(ob) = (L - 2*x ) * (  - 2*x ) * x

Wich is a function of x

V(x) = [( 12 - 2*x ) * ( 8 - 2*x ) ]*x

V(x) = ( 96 - 24*x - 16*x + 4*x²) * x

V(x) = 96*x - 40*x² + 4*x³

Tacking derivatives on both sides of the equation

V´(x) = 12*x² - 80*x + 96

V´(x) = 0     12*x² - 80*x + 96 = 0

Solving for x

x₁,₂ = 80 ± √ 6400 - 4608 / 24

x₁,₂ = 80 ± 42,33 / 24

x₁ = 5,10       We dismiss this solution since 2*x  becomes 2*5,10 = 10,20

ant this value is bigger than 8 inches

x₂ = 1,60 in

Therefore dimensions of the box

a = 12 - 2*x     ;  a =  12 - 3,20  ;  a = 8,8 in

b = 8 - 2*x   ; b = 8 - 3,20  ; b =  4,80  in

And the volume of the open box is:

V(max) = 8,8*4,8*1,6

V(max)  =  67,58 in³

How do we Know that is the maximun value for V?

We find  V´´(x)  = 24*x - 80     for  x = 1,6 is negative ( V´´(x) = - 41,6 therefore V (x) has  a local maximun for a value of x = 1,6