Calculus cardboard box problem I have a question regarding a rectangular box. The question is:

What are the dimensions of a rectangular box of greatest volume that can be constructed from 400 square inches of cardboard if the base is a square?Assume that the box has no top.

I know that that volume is length times height and width. But I am not sure how to set up a formula to solve the problem.
 A: Let's say the base is an $x \times x$ square, and call height $y$.
Since you know volume is length $\times$ width $\times$ height, we have that
$yx^2 = V\tag{1}$
You also know that you have $400$ square inches to work with.
The surface area of the box is $x^2$ (the area of the bottom of the box - no top since its to be an open box), plus $4\times x\times y$, the area of the 4 sides of the box.
So, $x^2 + 4xy = 400\tag{2}$
Now, we can use equatins $(1), (2)$ to put together a formula for maximizing Volume:
From $(2)$: $4xy = 400 - x^2 \iff y = \dfrac{400 - x^2}{4x}\tag{3}$
Substituting for $y$ from equation $(3)$ into equation $(1)$ gives us:
$$V = x^2y = x^2\left(\dfrac{400 - x^2}{4x}\right)$$
$$ \iff V = \frac{x(400-x^2)}{4} = \frac 14 x(400 - x^2) = 100x - \frac 14 x^3\tag{4}$$
Now, to maximize V, in terms of x, we differentiate $V$, set $V' = 0$, and determine critical points to test for maximums. 
In your derivative, you'll find you get a difference of squares, which factors fairly nicely, revealing the possible roots for $x$: one positive, one negative. So your maximum is going to have to occur when $x = $ positive root.
Once you've found $x_{\text{max}}$, use this to solve for $y$ (height), using equation $(3)$, and once you have $y$, you're ready to compute the maximum possible volume of the box (given the constraint of the total available cardboard) using equation $(1)$ and substituting into that equation the values you obtain for $x_{\text{max}}$ and $y$ to solve for V = volume, in cubic inches.
A: Steps


*

*Formulate the question mathematically: out of all the possible square-base, open-top boxes with surface area $400\,\text{in}^2$, we want to find the one with the greatest volume.

*Note that calculus provides a technique for finding the maximum of a function: we know that anywhere a (differentiable) function attains a maximum, its derivative must be zero there.

*Construct the formulas for the surface area and volume of a square-base, open-top box whose base has side length $b$ and whose height is $h$.

*Use the constraint that the surface area must equal $400\,\text{in}^2$ to express one of the variables in the surface area formula in terms of the other. Thus, each square-base, open-top box of surface area $400\,\text{in}^2$ is determined entirely by the value of first variable.

*Express the volume a of a square-base, open-top box of surface area $400\,\text{in}^2$ as a function of this one variable, take its derivative, and find where the derivative is 0. Then check by hand which of these point(s) are the maximum.
A: Hint: This problem is equivalent to maximizing $ V(x,y) = x^2y  $ given the constraint $x^2 + 4xy = 400$ (Note that the question specifies that box is open.) You can follow a similar procedure to amwhy's post now to solve the question.
