# Homework Help: Find the most economical dimensions of a closed rectangular box [. . .]

1. Mar 26, 2012

### s3a

"Find the most economical dimensions of a closed rectangular box [. . .]"

1. The problem statement, all variables and given/known data
Find the most economical dimensions of a closed rectangular box of volume 8 cubic units if the cost of the material per square unit for (i) the top and bottom is 5, (ii) the front and back is 2 and (iii) the other two sides is 5.

(a) What is the length of the vertical edge?
(b) What is the length of the horizontal front and back edge?
(c) What is the length of the horizontal side edge?

2. Relevant equations
Lagrange Multiplier Method or equivalent.

3. The attempt at a solution
The Lagrange Multiplier method is not required for this problem so if there is an easier way please notify me of it as well as why it's easier. My work is attached but I'm stuck at the step of figuring out the variables/dimensions.

Any help would be GREATLY appreciated!

#### Attached Files:

• ###### MyWork.jpg
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2. Mar 26, 2012

### HallsofIvy

Re: "Find the most economical dimensions of a closed rectangular box [. . .]"

It would make more sense if you would specify what x, y, and z represent at the start.
I assume you mean that x and y are "length" and "width" and z is "height".

In any case, yes, the volume satifies xyz= 8 and the "cost" is 10xy+ 4xz+ 10yz.

You can write each of your erquations as
$$10y+ 4z= -\lambda yz$$
$$10x+ 10z= -\lambda xz$$ and
$$4x+ 10z= -\lambda xy$$

Since the value of $\lambda$ is not important to the solution, I like to eliminate it first by dividing one equation by another:
If you divide the first equation by the second you get
$$\frac{5y+ 2z}{5x+ 5z}= \frac{y}{x}$$
and if you divide the first equation by the third you get
$$\frac{5x+ 2z}{2x+ 5z}= \frac{z}{x}$$
Those, together with $xyz= 8$ give you three equations to solve for x, y, and z.

3. Mar 26, 2012

### Ray Vickson

Re: "Find the most economical dimensions of a closed rectangular box [. . .]"