Optimization Problem:
A sheet of cardboard is $15$ inches wide by $25$ inches long and needs to be cut and folded into a rectangular base. The dark parts will be cut away and the light parts will be kept and folded including a $1$-inch flap at the top. What dimensions of the box would maximize the volume that the box could hold?
I have worked out the problem to the best of my knowledge. I left out the "flap" of the top because I thought it would be best to add it in the end. My work:
$V = lbh$:
$l = 24-2x$,
$b = \frac{15 - 2x}{2}=\frac{15}{2} - x$,
$h = x$
$\begin{align}V &= (24 - 2x)(\frac{15}{2} - x)x\\ &= x(180 - 24x - 15x + 2x^2)\\ &= 2x^3 - 39x^2 +180x\end{align}$
$\begin{align}\frac{dV}{dx} &= 6x^2 - 78x = 180\\
&= 6(x-3)(x-10) = 0\end{align}\\$
$\qquad\qquad\,{x=3}$, $x=10$
Plugging it into the original volume equation I got:
$V(10) = -100$ (not possible),
$V(3) = 243$ (maximum volume)
I am trying to compare my answers to make sure I am solving the problem correctly and get any advice to the way I solved the problem. I can't seem to figure out how to determine the actual lengths of the sides of the box so that I can figure out the area of the "flap".
Any help is greatly appreciated!
In the absence of a diagram that shows what the "dark parts" are, I interpreted the problem this way. Consider the figure below:
A square of side length $x$ is removed from each corner of the $25~\text{in} \times 15~\text{in}$ rectangle. The sides are then folded to form a box of length $(25 - 2x)~\text{in}$, width $(15 - 2x)~\text{in}$, and height $x~\text{in}$. In order to form a one inch flap at the top of each side, the sides of the box are folded over one inch from the top (the red lines in the diagram, so the flaps are the narrow rectangles between the red lines and the outer edges of the original box), leaving us with a box with length $(25 - 2x)~\text{in}$, width $(15 - 2x)~\text{in}$, and height $(x - 1)~\text{in}$. Thus, we have \begin{align*} V(x) & = (25 - 2x)(15 - 2x)(x - 1)\\ & = [25(15 - 2x) - 2x(15 - 2x)](x - 1)\\ & = (375 - 50x - 30x + 4x^2)(x - 1)\\ & = (4x^2 - 80x + 375)(x - 1)\\ & = x(4x^2 - 80x + 375) - 1(4x^2 - 80x + 375)\\ & = 4x^3 - 80x^2 + 375x - 4x^2 + 80x - 375\\ & = 4x^3 - 84x^2 + 455x - 375 \end{align*} Differentiation yields $$V'(x) = 12x^2 - 168x + 455$$ Setting the derivative equal to zero yields \begin{align*} x & = \frac{-(-168) \pm \sqrt{(-168)^2 - 4 \cdot 12 \cdot 455}}{2 \cdot 12}\\ & = \frac{168 \pm \sqrt{28224 - 21840}}{24}\\ & = \frac{168 \pm \sqrt{6384}}{24}\\ & = \frac{168 \pm 4\sqrt{399}}{24}\\ & = \frac{42 \pm \sqrt{399}}{6}\\ & = 7 \pm \frac{\sqrt{399}}{6} \end{align*} Observe that $$\frac{\sqrt{399}}{6} \approx \frac{20}{6} = \frac{10}{3} \implies 7 + \frac{\sqrt{399}}{6} > \frac{15}{2} \implies 15 - 2x < 0 $$ so we can immediately discard the positive root on physical grounds.
If we apply the First Derivative Test, we see that $V'(x)$ changes from positive to negative at the critical point $x = 7 - \frac{\sqrt{399}}{6}$, so we may conclude that the volume has a relative maximum at the critical point. Moreover, the volume is $0$ when $x = 0$ or $x = 15/2$. Hence, the maximum volume occurs when $$x = 7 - \frac{\sqrt{399}}{6} \approx 3.67$$
If you prefer the Second Derivative Test, observe that $$f''(x) = 24x - 168$$ is negative when $x < 7$, which also leads to the conclusion that $x = 7 - \frac{\sqrt{399}}{6}$ is a relative maximum.
The corresponding dimensions for the box are \begin{align*} l & = (25 - 2x)~\text{in}~\approx 17.66~\text{in}\\ w & = (15 - 2x)~\text{in}~\approx 7.66~\text{in}\\ h & = (x - 1)~\text{in}~\approx 2.67~\text{in} \end{align*} which yields a volume of approximately $361.19~\text{in}^3$.