# optimatization problem

• Feb 9th 2009, 09:35 PM
skorpiox
word problem
Hi! can someome help me with this exercise

An open top box with a rectagular base is to be constructed. The box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. What should the dimensions of the box be if the surface area is to be
a. 90 square inches ? b. as small as possible?

I have already done some of these problems , but I don't know how to set up this one.
• Feb 9th 2009, 10:11 PM
mollymcf2009
Quote:

Originally Posted by skorpiox
Hi! can someome help me with this exercise

An open top box with a rectagular base is to be constructed. The box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. What should the dimensions of the box be if the surface are is to be
a. 90 square inches ? b. as small as possible?

I have already done some optimatization problems , but I don't know how to set up this one.

I think that you have your wording wrong in your problem. Are you sure it doesn't say something like, "twice as long as it is tall"?
Check that, but here is how you need to set this up:

Volume = length x width x height

Surface area (box with no top) = 2(length x height) + 2(width x height) + (length x width)

V = 8h
A = 2(4h) + 2(2h) + 8
= 8h + 4h +8
= 12h + 8

Double check the question and then I can help you through the rest. IT really isn't an optimization problem the way it's worded.

Since you are given your volume, you should have two variables in there.
• Feb 9th 2009, 10:31 PM
skorpiox
Quote:

Originally Posted by mollymcf2009
I think that you have your wording wrong in your problem. Are you sure it doesn't say something like, "twice as long as it is tall"?
Check that, but here is how you need to set this up:

Volume = length x width x height

Surface area (box with no top) = 2(length x height) + 2(width x height) + (length x width)

V = 8h
A = 2(4h) + 2(2h) + 8
= 8h + 4h +8
= 12h + 8

Double check the question and then I can help you through the rest. IT really isn't an optimization problem the way it's worded.

Since you are given your volume, you should have two variables in there.

I double checked the problem , and it says
an open-top box with a rectangular base is to be constructed. the box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. what should the dimensions of the box be if the surface area is to be
a. 90 square inches? b.as small as possible?
• Feb 9th 2009, 11:06 PM
mollymcf2009
Ok, look at it this way:

Let's call the width (x), so the length is (2x), height we will call (h)

Volume of the box is:$\displaystyle x \cdot 2x \cdot h$ = $\displaystyle 2x^2h$

Now, your surface area is: $\displaystyle 2xh + 4xh + 2x^2$
**That is: $\displaystyle 2(width\cdot height) + 2(length \cdot height) + (length \cdot width)$ 2 short sides + 2 long sides + bottom of box

So,

$\displaystyle V = 2x^2h$
$\displaystyle SA = 6xh + 2x^2$

Using these two equations and the data given to you to figure out the dimensions of your box.

To find the smallest area, take the area equation we came up with (after you plug in your height you solved for), take the derivative and then set it equal to zero. Solve for x. This is your critical value. Plug it into the original area function and solve for y. This will give you your minimum value (you can do the same thing to find a maximum value too!. Hope that helps you!