Another extrema problem

[sigma]

A rectangular container, with open top, is required to have a volume of 16 cubic meters. Also, one side of the rectangular base is required to be 4 meters long. If material for the base costs $8 per square meter, while material for the sides costs $2 per square meter, find the dimensions of the container so that cost of material to make it will be a minimum.

I'm sorry I don't have any work. Don't quite know where to begin. The one side of the base being 4 meters is throwing me off. What I can tell you though is the answer is:\(\displaystyle \
\L\
4\) X\(\displaystyle \
\L\
sqrt 2
\\) X \(\displaystyle \L\2\sqrt 2\)

 

[tkhunny]

A rectangular container, with open top, is required to have a volume of 16 cubic meters. Also, one side of the rectangular base is required to be 4 meters long. If material for the base costs $8 per square meter, while material for the sides costs $2 per square meter, find the dimensions of the container so that cost of material to make it will be a minimum.

I'm sorry I don't have any work. Don't quite know where to begin. The one side of the base being 4 meters is throwing me off. What I can tell you though is the answer is:\(\displaystyle \
\L\
4\) X\(\displaystyle \
\L\
sqrt 2
\\) X \(\displaystyle \L\2\sqrt 2\)

By the time you get to calculus, you are not supposed to be stumped by the geometry. Make a drawing and label the parts.

The base is 4 * Something -- How about 4 * W? Length = 4, Width = W
The height, we also don't know. Let's call that H.

We are given: 4 * W * H = 16

There is no top, so only one "Base" 4 * W
Two sides of one size: 2 * 4 * H
Two sides of the other size: 2 * W * H

How much does it cost to build a box?

Cost(W,H) = 4*W*8 + (2*4*H + 2*W*H)*2

Simplify this

Cost(W,H) = 32*W + 16*H + 4*W*H

We COULD realize that minimizing Cost(W,H) will produce the same result as minimizing Cost(H,W)/4, but I'll skip over that. The real problem for now is that this is still in two variables. Make it only one variable by substituting from the given information about the volume.

4 * W * H = 16 ==> H = 16/(4*W) = 4/W

Cost(W,H) = 32*W + 16*H + 4*W*H
Cost(W) = 32*W + 16*(4/W) + 4*W*(4/W)

OK. You should be able to minimize that. Find dCost(W)/dW.

 

[sigma]

Great thanks. I got the answer. It actually turned out to be a really easy question in the end with a simple derivative. Its not that I'm stumped by geometery, but it is one of my weak areas. The hardest part to these problems is setting up the equation for what you want to be optimized, so you have to know the proper relations with the various shapes they give. I also wasn't sure if it was area or surface area that was to be optmized. And with the 4 thing, I just wasn't sure if the 4 should replace the W, H or L but I guess it doesn't matter. In my calclus class, they usually use X and Y in leau of L and W but this almost makes more sense to me. Once I drew out the 3D rectangle, I could see the relations easily. Here's my solution if anybody is interested.

\(\displaystyle \
\L\
\begin{array}{l}
\frac{{dc}}{{dx}} = 32 - \frac{{64}}{{w^2 }} \\
\\
0 = \frac{{32w^2 - 64}}{{w^2 }} \\
\\
0 = 32(w^2 - 2) \\
\\
w = \pm \sqrt 2 {\rm but not } - \sqrt 2 \\
\\
h = \frac{4}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{4\sqrt 2 }}{2} = 2\sqrt 2 \\
\\
{\rm the dimentions are }4{\rm x }2\sqrt 2 {\rm x }\sqrt 2 \\
\end{array}
\\)

 

[tkhunny]

Since I am on a mission for pre-calculus mathematics this week, let me first pat you on the back for getting the idea, but second, let me encourage you to upgrade your geometry so that you no longer have it as a weak area.