A box with no top must have a base twice as long as it is wide, and the total surface area of the box is to be 54 ft^2. What is the maximum possible volume of such a box?
Width=X
Base=2*X
Height=Y
Surface area w/o a top is Base*Width + 2*Base*Height + 2*Width*Height
54 = 2X*X + 2*2X*Y+2*X*Y = 2X^2+6X*Y
Solve for Y:
54-6*X*Y=2*X^2
-6*X*Y=2*X^2-54
Y=(2*X^2-54)/(-6*X)
Equation for volume is Base*Width*Height
We now know what Y is in terms of X so we just multiply them all together.
X*2X*(2*X^2-54)/(-6*X)=18*X^3-(2/3)*x^5
Differentiate in terms of X so we can set equal to zero to find maximum.
d(B*W*H)/dx=54*x^2-(10/3)*X^4
54*x^2-(10/3)*X^4 = 0 for x = 9*(5)^(1/2)/5 or approximately 4.02492
Sorry, to find maximum volume, plug that value into our volume formula..
18*(4.02492)^3-(2/3)*(4.02492)^5 = 469.467 ft^3