An excellent problem. What is your plan for solving it?
Did you draw a box?
If so, how big is the base? How big are the sides? Are the sides square, too? What is the volume of this general box?
You CANNOT have NO idea. Let's see what you get.
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 .