# Thread: Minimum Material Dimensions of a Box

1. ## Minimum Material Dimensions of a Box

A company wishes to manufacture a box with a volume of 36 ft^3 that is open on top and is twice as long as it is wide. Find the dimensions of the box produced from the minimum amount of material.

I must have missed class the day we went over this. I have no clue how this is supposed to be done. Please help!

2. width = $x$

length = $2x$

height = $h$

Now $2x\times x \times h = 36$

All you need now is a relation ship that links $h$ and $x$

3. Yeah. That is the part I am having trouble with. There seems to be no example in the book that it relates to either. I now have 2x^2(h) = 36

4. How about $h = \frac{36}{2x^2} = \frac{18}{x^2}$ ?

Next step is to think about the problem.

You want to minimise the materials (the surface area) of the 5 sided box.

Find an expression for the surface area is terms of $x$ and $h$

Spoiler:

$SA= 2\times 2xh +2\times xh+2x^2$

Do you know why?

$SA = 2\times 2x\times \frac{18}{x^2} +2\times x\times \frac{18}{x^2}+2x^2$

Now to finish solve

$\left( 2\times 2x\times \frac{18}{x^2} +2\times x\times \frac{18}{x^2}+2x^2\right)' = 0$

5. Sorry, I'm not really understanding this. My question asks me to "write an equation for the quantity you are supposed to minimize," and also to "find a function for what I'm supposed to minimize" too. So would...h = 18/x^2 be the equation and then the function would be what you wrote with the surface area?

6. Originally Posted by trouty323
Sorry, I'm not really understanding this. My question asks me to "write an equation for the quantity you are supposed to minimize," and also to "find a function for what I'm supposed to minimize" too. So would...h = 18/x^2 be the equation and then the function would be what you wrote with the surface area?
"write an equation for the quantity you are supposed to minimize" and "find a function for what I'm supposed to minimize" are referring to the same thing- the "quantity" and "function" are the same- here the "quantity" is the "minimum amount of material" which is surface area. No, "h= 18/x^2" is NOT "the equation for the quantity you are supposed to minimize". The "equation" is "surface area= base area+ 2 side area1+ 2 side area2. If you take the length of the base to be "x", the width of the base to be "y", and the height to be "h", then because the box is to be "twice as long as it is wide", we have y= 2x.

The base area is $xy= x(2x)= 2x^2$. The area of each side on the side with width x is xz and the area of each side on the side with width y is yz= 2xz so the total surface area is $A= 2x^2+ 2(xz)+ 2(2xz)= 2x^2+ 6xz$.

Finally, you are told that the volume is to be 36 cubic feet: the volume is given by $xyz= x^2z= 36$ so $z= \frac{36}{x^2}$. Putting that into the equation for A, we have $A= 2x^2+ 6x(\frac{36}{x^2}= 2x^2+ \frac{216}{x}$. That is the "equation for the quantity you are supposed to minimize" and the function is $2x^2+ \frac{216}{x}$.