Maximum and Minimum Problem - What are the dimensions of the open box that meets the following specifications?

Question

What are the dimensions of the open box that meets the following specifications? a) has maximum volume b) the base is a rectangle twice as long $(l)$ as it is wide $(w)$ c) you have $2400$ square inches of material to use (no waste)

I think I have to find the derivative of $v=whl$ and set it to $0$ but other than that I am unsure of what to do.

Answer 1

Set the width of the box is $a$. Then, its length is $2a$. Suppose the box is $h$ tall (all measures are in $inches$). Then it has $2a^2$ base, and $2 \times ha$ $2 \times 2ha$ sides, all in $inch^2$. Thus, total surface area is $2400 inch^2$ such as: $$2400=2a^2+2\left(ha+2ha \right)$$ Thus, $3ha=1200-a^2$ and hence, $h=\frac{1200-a^2}{3a}$ Now, volume of the box is $V$, then $$V=(a)(2a)(h)=2a^2\left(\frac{1200-a^2}{3a}\right)=2a\left(\frac{1200-a^2}{3}\right)$$ Now, get the first derivative of $V$ with respect to $a$: $$\frac{dV}{da}=2\left(\frac{1200-a^2}{3}\right)+2a\left(\frac{-2a}{3}\right)=2\left(\frac{1200}{3}-\frac{3a^2}{3}\right)=2\left(400-a^2\right)$$ For critical points:$\frac{dV}{da}=0$, thus, $400-a^2=0$ $\Rightarrow a=\pm20$

Now, get the second derivative of $V$ with respect to $a$: $$\frac{d^2V}{da^2}=2\left(-a\right)=-2a$$ $\frac{d^2V}{da^2}\lt 0$ only when $a=+20$ and that is maximum. Thus, the maximum volume of the box,$V_{max}$: $$V_{max}=2\times20\left(\frac{1200-20^2}{3}\right)=10666\frac{2}{3} inch^3$$

Answer 2

I think I have to find the derivative of $v=whl$ and set it to $0$ but other than that I am unsure of what to do."

Close. In order to do this, you need a function of one variable. So, right now, your function is in three variables, $w$, $h$, and $l$. You need more equations to reduce them. Thankfully, there is information in your problem that can add additional equations to reduce the number of variables!

1) "$2400$ square inches of material to use" - that's your surface area. Do you know of the equation for the surface area of a box?

2) "the base is a rectangle twice as long as it is wide" - that is an equation that relates two of your sides.

You can use #$1$ and #$2$ to reduce the problem down to where $v$ is in terms of a single variable. Then take the derivative and set it to zero. The location where this is found is the value that has the maximum.

Then, go back and use #$1$ and #$2$ again to solve for the other dimensions.

Technically, you should also check the second derivative to make sure that this is a maxima and not a minima, and check the graph to make sure that there aren't any other maxima and minima you are missing. But, given the problem, I'm sure your answer will be the maximum.