Question 1119872
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Let the side of the square base be x and the height be h.  Then<br>
(1) the total surface area (base and 4 sides) in square meters is 432:
{{{x^2+4xh = 432}}}<br>
(2) the volume (to be maximized) is the area of the base times the height:
{{{V = x^2h}}}<br>
Solve equation (1) for h and substitute in equation (2) to get an expression for the volume in terms of the single variable x:<br>
{{{4xh = 432-x^2}}}
{{{h = (432-x^2)/4x}}}
{{{V = x^2((432-x^2)/4x) = (432x^2-x^4)/4x = 108x-x^3/4}}}<br>
Take the derivative and set it equal to 0 to find the value of x that maximizes the volume; and calculate the volume for that value of x:<br>
{{{108 - (3/4)x^2 = 0}}}
{{{(3/4)x^2 = 108}}}
{{{x^2 = 144}}}
{{{x = 12}}}
{{{V = 108(12)-(12^3)/4 = 1296-432 = 864}}}<br>
The maximum volume is 864 cubic meters, when the square base is 12m on a side and the height is 6m.