Let $x$ be the length of a side of the square end, and let $y$ be the length of the box. As you say in your comment, the volume is given by $V=x^2y$, and the postal service’s restriction limits $4x+y$ to a maximum of $108$. Clearly we should use the entire allowance, so we want $4x+y=108$. However, you’ve solved this incorrectly for $y$: $y=108-4x$, not $\frac{108}{4x}$. Substituting the correct expression for $y$ into the volume formula, we get $V=x^2(108-4x)=108x^2-4x^3$.
You want to find the value of $x$ that maximizes $V$, knowing that $V=108x^2-4x^3$. How can you use the first derivative $\frac{dV}{dx}$ to find this value of $x$? (I’ll stop here to give you a chance to think about it; if you get completely stuck, leave a comment.)
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.
Best Answer
Let us say we have a box. We know that the volume is $l \cdot w \cdot h$.We know that we are trying to optimize this problem with the constraint that $2 w+2 h+l$ is $112$. We know that the base is a square so the volume is now $l^2 h$. We also happen to know that $w=h$. So we have $4w+l=112$. We can isolate the constraint as such: $l=112-4w$. We then get this cubic $(112-4w)^2 \cdot w$. I am sure that you can do the rest.