If 2100cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Answer is in cm^3
haha ok well I tired i million different ways to do this problem and nothing is right. but this is how I thought the right way to do it was
since we have 2100cm^2 to work with, that would be equal to suerface area. We don't know if any of the other sides are squares, just that the base is a square, so the surface area = x^2+2xy = 2100cm^2
I solve for y to get y = (2100-x^2)/2x
I plugged that into the volume which would be V= (x^2)(y)
so that would be V = [x^2][(2100-x^2)/2x]
then i found the derivative which is
V' = [(2x)((2100-x^2)/2x)] + [((x^2)(-2x)(2x)-(2100-x^2)(2))/(2x)^2]
and then when I try to solve for x i just end up getting 0, so I know I'm doing this problem wrong.
I got it right!! Thanks!
SA = x^2 + 4xy = 2100 cm^2
I solved for y to get y = (2100-x^2)/4x
Plugged y into V = (x^2)(y)
which gave me V = [(2100x^2)-(x^4)]/4x
then found the derivative which gave me
V' = [(4200x - 4x^3)(4x)-(2100x^2 - x^4)(4)]/(4x)^2
I simplified that and got
V' = [4x^2(4200 - 4x^2) - (2100 - x^2)]/(4x)^2
then I solve for x to get x = sqrt 700
I plugged x into SA = x^2 + 4xy = 2100 to solve for y
Then I plugged both x and y into V = (x^2)(y) to get
V = 9260.129589cm^2 which is the correct answer!
A point, I mentioned that you should try to simplify as much as possible before you differentiate. While what you did is completely correct, it is a lot of work. You differentiated V = [(2100x^2)-(x^4)]/4x, to get V' = [4x^2(4200 - 4x^2) - (2100 - x^2)]/(4x)^2. Simplify V first and you get V= (2100x-x^3)/4, which gives V' = (2100-3x^2)/4.
Much easier. Work smart, not hard .