Question

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bending upward the projecting portions which remain. Find the maximum volume that can be obtained.

Answer 1

Answer:

V (max)  = 128 in³

Step-by-step explanation:

Let call " x " the length of the side of the square cut from the cornes, then

each side of the square base of the box would be

y = 12 - 2x

The volume of the future box is:

V = ( 12 - 2x ) * ( 12 - 2x ) * x

V(x) = ( 144 -24x - 24x + 4x² )* x

V(x) = 4x³ - 48x² + 144x

Taking derivatives  on both sides of the equation we get

V´(x) = 12x² - 96x + 144

V´(x) = 0   ⇒    12x² - 96x + 144 = 0    ⇒   x² - 8x + 12  = 0

x² - 8x + 12  = 0     Solving this second degree equation

x₁,₂  = [ 8  ±  √ 64 - 48 ] /2

x₁   =  [ 8 + 4 ] / 2      ⇒    x₁   =  6  we dismiss this solution since it means non base at all

x₂  =  [ 8 - 4 ] / 2      ⇒ x₂  = 2

If we try this value in the second derivative we will find

V´´(x) =  2x - 8    ⇒   V´´(x)  < 0  that means there is a maximun in this point

Then the base of the box is

y  =  12 - 2x     ⇒    y  =  8 in

and the height    x  =  2 in   and

V (max)  =  8*8*2

V (max)  = 128 in³