Thread: maximal surface area of a box

1. maximal surface area of a box

I'm trying to figure out the second part of this problem. I would appreciate any help.

A closed rectangular box with a square base is to be built subject to the following conditions: the volume is to be 27 cubic feet, the area of the base may not exceed 18 square feet, the height of the box may not exceed 4 feet. Determine the dimensions of the box (a) for minimal surface area; (b) for maximal surface area.
(a) The minimal surface area is sq. ft. at the side length of the square base = ft.
(b) The maximal surface area is approximately sq. ft. (round your answer to 2 decimal places) at the side length of the square base =

I was able to solve for the minimal surface area, but i'm having trouble finding the maximum.

2. Maxima and minima are only found at 'critical points' of continuous functions. Critical points are found at:

1. Endpoints of the domain,
2. Points at which $\displaystyle y'=0$, and
3. Points at which $\displaystyle y'$ doesn't exist.

Here, we can set up our equations as follows:

\displaystyle \begin{aligned} S&=2b^2+4bh\;\;\;\;\;\;\;\;\;\mbox{area of top and bottom + area of four sides}\\ A&=b^2h=27. \end{aligned}

Our task is first to find the critical points of $\displaystyle S$. Hint: $\displaystyle h$ may be expressed in terms of $\displaystyle b$.

3. ok, so after finding the derivative, I got b=0 at 3. This critical point is the local minimum. finding f(3), i got
$\displaystyle 54ft^2$

I'm just not sure how to set this up to find the maximum surface area.

4. Originally Posted by discobob

I'm just not sure how to set this up to find the maximum surface area.
You can express the surface area in terms of b and calculate the derivatives:

$\displaystyle S(b)=2b^2+108/b$

$\displaystyle S'(b)=4b-108/b^2$

$\displaystyle S''(b)=4+\frac{2(108)}{b^3}$

Note both derivatives tell us the extremum is a minimum. Now, you're subject to the constraints $\displaystyle b\leq \sqrt{18}$ and $\displaystyle h\leq 4$. But $\displaystyle h=\frac{27}{b^2}$ so $\displaystyle \frac{27}{b^2}\leq 4$. That means $\displaystyle b\geq \sqrt{27/4}$. That means you want to find the maximum surface area in the interval $\displaystyle \sqrt{27/4}\leq b \leq \sqrt{18}$. But by the derivatives above, the function has a minimum at b=3 and is concave upwards on both sides of this point. Then where must the maximum of this function lie in this interval then?

5. oh, ok, i get it, the maximum surface area is infinity. Thanks for the help

6. Originally Posted by discobob
oh, ok, i get it, the maximum surface area is infinity. Thanks for the help

No. If $\displaystyle S(b)=2b^2+108/b$ and $\displaystyle S(b)$ is continuous and concave up in the interval $\displaystyle \sqrt{27/4}\leq b \leq \sqrt{18}$, then the maximum of $\displaystyle S(b)$ in that interval occurs at the end points. So:

$\displaystyle S(\sqrt{27/4})\approx 55$

$\displaystyle S(\sqrt{18})\approx 61.46$

Draw a picture of $\displaystyle S(b)$ and convince yourself of this fact.