Question 218034
Dalton has a closed rectangular box with a square base at least 10 inches long
 have a surface area of 900 square inches.
 what dimensions will give the box a maximum volume?
;
Let x = the side of the square base
Actually the maximum is when the box is a perfect cube, therefore
let x = the height of the box also
:
Surface area equation, find h:
6x^2 = 900
simplify divide by 6
x^2 = 150
x = {{{sqrt(150)}}}
x = 12.247"
:
Max vol: 12.247 by 12.247 by 12.247, and 12.247^3 = 1836.9 cu/in
:
You can prove this: 
let the height = h, 
:
Surface area equation, find h:
2x^2 + 4xh = 900
simplify divide by 2
x^2 + 2xh = 450
2xh = -x^2 + 450
h = {{{(-x^2 + 450)/(2x)}}}
:
Volume equation
V = x^2 * h
Replace h
V = x^2({{{(-x^2 + 450)/(2x)}}})
V = x({{{(-x^2 + 450)/2}}})
V = x({{{(-x^3 + 450x)/2}}})
 
Graph this equation y=V
{{{ graph( 300, 200, -6, 30, -1000, 2000, (-x^3+450x)/2) }}}
:
Shows max volume when x is a little over 12"; vol is about 1800 cu/in