Problem with solving an equation

[Parashurama]

Homework Statement

I have a square 50x50 units. I want to make a pool(box without a lid), and then cut away the corners. The "corner-square" is x^2.
What I want is a formula X(v) that gives me the x value needed to get a fixed volume.

Homework Equations

The V(x) function is base x height, x(50-2x)^2. But i have problems isolating x based on this formula.

The Attempt at a Solution

If i try solving it in Maple I get:
25/2 + 1/2sqrt(625-2v) and 25/2 - 1/2sqrt(625-2v)
but this makes no sense to me.

I've calculated the max volume at x=25/3 to be 250000/27, and the function is only valid from x=0..25.

I guess the answers Maple gives me is because it does not know og my 0..25 range, but I don't know how to make that part of the calculation.

If the problem is unclear, please say so and I will try to explain better.

 

[HallsofIvy]

If each side is 50 and you "cut away" $x^2$ from each corner then you base will have sides of length 50- 2x. The base will have area $(50- 2x)^2$ and the box will have height x. The volume is $x(50- 2x)^2$.

Now, what is it you specifically want to do? Solve $x(50- 2x)^2= v$ for any v? Multiplying out the left side gives $4x^3- 200x^2+ 2500x= v$ or the cubic equation $4x^2- 200x^2+ 2500x- v= 0$. Your "Maple" solution looks like a quadratic formula- you may have left out an "x".

 

[Parashurama]

Thanks for answering!
What I want is a function that gives me the x value needed to make a box with a cirtain volume.

I will check my equations in Maple tonight, when I get back from work.

 

[Parashurama]

I did leave out an x, but still did not get a sensible answer. But when I read the problem text once more a saw that I was not supposed to make a formula, but use the formula V=x(50-2x)^2 and use tables to narrow down to the volume you needed.

So the problem is solved that way and that I figuerd out :P

Thanks for the help!