Help: Trouble thinking through an optimization prob. (Math)

Soontir C'boath · Post by **Soontir C'boath** » 2005-12-01 07:47pm

Let me start off with the question:

A container with square base, vertical sides, and open top is to be made from 1000 ft^2 of material. Find the dimensions of the container with greatest volume.

I want to know if I got this right.
A = L^2 + 4(L*H) (Which accounts for the square base and the four sides)

Then we derive it with respect to L which gives us: A' = 2L + 4H
Next we set it to equal to zero which then allows us to see that the height H = L/2

Substituting H now with L/2 into the area formula should then give us the dimensions of the object.

My questions are, aren't I suppose to do something with the volume and is what I did above correct? Is it right except that I should calculate what the volume is for what I got above and compare it to say...a cube to make sure it is the greatest?

Exonerate · Post by **Exonerate** » 2005-12-01 09:11pm

It's been awhile since I did this so I might've forgotten something...

V = L^2 * H
(Shouldn't H = -L/2 ?)
Substitute the H in, then derive the volume with respect to L. Then find the second derivative and factor to find the extremas. I'm probably wrong here, but I'm fairly sure the answer is along this line of reasoning...

Soontir C'boath · Post by **Soontir C'boath** » 2005-12-01 09:17pm

Yea the negative stumped me because lengths shouldn't have any negatives!

Soontir C'boath · Post by **Soontir C'boath** » 2005-12-01 09:29pm

Here's a different approach I took.

Instead of deriving the area I kept it the same and solved for H:

H = .25[(1000/L) - L]

Then substituted into V:

V = L^2 * .25[(1000/L) - L] = .25 (1000L - L^3)

Then V' = .25 (1000 - 3L^2) and solve for zero (The .25 disapears when we cross multiply)

So then:

L = sqrt(1000/3) = 18.26

And H = 9.13

Well! Looks like I don't need any help at all. Well, I'm definitely relieved. My brain is still working!

fnord · Post by **fnord** » 2005-12-01 09:32pm

Well, atempting to tie together what Exonerate did with what you've already done., taking L as the side length of the base and H as the height.

Volume of the box V = L^2 * H

The constraint we have to work with is a cap on the area: A = L^2 + 4LH

The maximum area you've given is 1000 sq ft, so: 1000 = L^2 + 4LH
(1000 - L^2)/4L = H (rearranging)

We then substitute the expression for H into V

V = L^2 * (1000 - L^2)/4L
= L * (1000 - L^2)/4
= 250L - L^3 / 4

Differentiating to find the extrema

dV/dL = 250 - 3L^2/4

0 = 1000 - 3L^2

L = +/- sqrt(1000/3) We can discard the negative solution, as length is nonnegative.

L = sqrt(1000/3)

L is constrained by the problem to lie in the closed interval [0,sqrt(1000)] (base of zero side length at one extreme, box of zero height at the other).

To check our results, find the second derivative.

d/dL (dV/dL) = -6L

OK, we have a negative second derivative everywhere in the interval bar zero, so the stationary point found earlier must be a maximum.

L = 0, V = 0
L = sqrt(1000), V = sqrt(1000)^2*0 = 0

L = sqrt(1000/3), V = 1000/3 * (1000 - 1000/3) / sqrt(1000/3)
V = 1000/3 * 2000/3 * sqrt(0.003)
= 12,171.61 cu ft (to 2 dec pl)

Soontir C'boath · Post by **Soontir C'boath** » 2005-12-01 09:39pm

Just to refresh myself, the second derivative is to check whether it's a max or min but not necessarily the global max or min.?

nickolay1 · Post by **nickolay1** » 2005-12-01 10:14pm

Yes. If it's negative, it's a max (the slope of the function starts out positive, becomes 0, then becomes negative; this implies that it was growing, then stopped, then began to decline, i.e. reached a maximum where the first derivative was 0). The opposite applies for a positive second derivative. If the second derivative is 0, it means that the rate of change of the function at the point is 0, but that the function did not change direction (decreasing or increasing).

Wyrm · Post by **Wyrm** » 2005-12-02 08:23pm

Some general cautions on max/min problems:

(a) The derivitive method only works where the derivitive exists. Therefore, the function must be continuous and differentiable in some small neighborhood in the vicinity of the max or min. If not, you will have to find out where the function is discontinuous, find its value there, and compare it to other extrema you find.

(b) Check the edges of the domain. If the domain is open in any way, then you could easily find yourself in a situation where the function has no global extrema. An example is the function f(x) = x on the domain 0<x<1. f goes to a limit of 1 as x approaches 1 and a limit of 0 as x approches 0, but for any x in the domain, you can always find a y and z such that f(y) < f(x) < f(z). Thus, f has no global maximum or minimum (there are always numbers between any given x and 0, and any given x and 1). On compact domains and functions that are continuous everywhere, this will not occur.

(c) Functions will sometimes have multiple extrema. Fortunately, for nice functions they will be finite in number. From there, it's just a matter of plugging in the numbers and comparing the outputs.