A container with square base, vertical sides, and open top is to be made from 200 ft^2 of material. Find the dimensions (height and length of base) of the container with greatest volume. If appropriate, leave your answer in radical form and enter all fractions in lowest terms.

as I told you six hours ago -- did you even bother to look?

If the base has sides x, and the height is h, then the area is
x^2 + 4xh = 200
so, h = (200-x^2)/(4x) = 50/x - x/4
Now the volume is
v = x^2 h = x^2 (50/x - x/4) = 50x - x^3/4
Now just find where dv/dx = 0 for maximum volume.
Post your work if you get stuck

I did check. It just closed the window and I didn’t remember exactly what you said and I couldn’t find the post so I reposted. Sorry.

As for my work, I tried solving for dv/dx and I get 50 - 3/(4(x^1/4)) and when I plug 0 in for that I get 81/1600000000. But how would I use this in order to find the specifics involving the length for the base of the box and the height? You show the formula for the height but is this number x or?

Really? You need to read more carefully ...

that's not x^(3/4) It is (x^3)/4
dv/dx = 50 - 3/4 x^2

Okay that makes much more sense. Sorry for misreading. Thanks for the help. I figured it out now.

To find the dimensions of the container with the greatest volume, we need to use optimization techniques. Let's break down the problem into smaller steps:

Step 1: Define the variables
Let's assume that the dimensions of the square base are both x units, and the height of the container is h units.

Step 2: Set up the equations
We know that the container has a square base, so its area is x * x = x^2 square units.
The four sides of the container have an area of 4 * x * h = 4xh square units.
The total surface area of the container is the sum of the base area and the side area:
x^2 + 4xh = 200 square units.

Step 3: Solve for h
Now, we can solve the equation for h in terms of x. Rearrange the equation:
4xh = 200 - x^2,
h = (200 - x^2) / (4x).

Step 4: Express the volume in terms of x
The volume of the container is given by V = x^2 * h. Substitute the expression for h:
V = x^2 * (200 - x^2) / (4x).
Simplify and rewrite the expression:
V = (200x - x^3) / 4.

Step 5: Find the critical points
To find the critical points, take the derivative of V with respect to x:
dV/dx = (200 - 3x^2) / 4.
Set this derivative equal to zero and solve for x:
200 - 3x^2 = 0,
3x^2 = 200,
x^2 = 200 / 3,
x = √(200 / 3).

Step 6: Determine the dimensions
We need to check whether this critical point corresponds to a maximum or minimum volume. To do this, we can take the second derivative of V with respect to x:
d²V/dx² = -6x / 4 = -3x / 2.

Since the second derivative is negative for x = √(200 / 3), this critical point corresponds to a maximum volume.

Therefore, the dimensions of the container with the greatest volume are:
Base length = x = √(200 / 3),
Height = h = (200 - x^2) / (4x).