```Question 236021
The bottom of the box and all four sides are constructed of glass costing \$1/ft2,
and
the top is to be constructed of glass costing \$5/ft2.
Sally has \$72 to spend on the box.
How should she choose the dimensions so that the volume of the box is greatest?
:
Let x = the side of the square base and top
Let h = height of the box
5 of the sides cost \$1 a sq/ft
Top cost 5x^2
:
We need to get h in terms of x
:
Top + bottom + 4 sides = 72
5x^2 + x^2 + 4(xh) = 72
6x^2 + 4(xh) = 72
4xh = 72 - 6x^2
h = {{{(72-6x^2)/4x}}}
Simplify
h = {{{(36-3x^2)/(2x)}}}
:
Find the volume:
V = x^2*h
replace h
V = x^2*{{{((36-3x^2))/(2x)}}}
Cancel x
V = x*{{{((36-3x^2))/2}}}
V = {{{((36x-3x^3))/2}}}
Write the equation as:
-1.5x^3 + 18x = 0
:
Graph this
{{{ graph( 300, 200, -2, 5, -10, 30, -1.5x^3+18x) }}}
Max volume when x=2
:
find h, when x=2
h = {{{(36-3(2^2))/(2*2)}}}
h = {{{(36-12)/(4)}}}
h = {{{24/4}}}
h = 6
;
A box: 2 * 2 * 6 = 24 cu (as seen on the graph)
;
Check the cost;
Top + bottom + 4 sides = \$72
5(4) + 4 + 4(2*6) =
20 + 4 + 48 = \$72```