A closed rectangular box with square ends has a total surface area of 2400 cm^2. Find the greatest volume the box can contain. Prove it so.

L = W (Square Ends)

H*L^2 = Volume (V)

2*L^2 + 4*L*H = 2400 cm^2

Lets Make it Look more like a quadratic because we can find max/min values with quadratics

Subtract the 2400

2*L^2 + 4*L*H - 2400 = 0

Problem is we don't need to find a maximum value for H, we need volume... lets rearrange another equation

H = V/L^2

Plug it in to get rid of the H in our quadratic

2*L^2 + 4*L*(V/L^2) - 2400 = 0

2*L^2 + (4*L*V)/L^2 - 2400 = 0

L cancels

2*L^2 + (4*V)/L -2400 = 0

We need volume on the right so lets move the whole term over

2*L^2 - 2400 = 0 - (4*V)/L

2*L^2 - 2400 = -4V/L

In order to get V by itself we need to get -4/L to equal one because 1*V = V. To do this we multiply by an inverse. Just remember that what you do to one side of the equation you must do to the other.

L/-4(2*L^2 - 2400) = -4LV/-4L

(-2/4)L^3 + 600L = V

If you graph it you get a standard inverted s-shape. This supposes that it goes on forever, however, if you think about it L can't be negative, so there is a definite maximum. You can use tables between 15 and 25 to find a maximum or if you like limits or even using derivatives. You can find the Value of L and V using the above equations but if you plug those numbers into several other equations you can find the values of the other letters. To check your work plug your final numbers into the Area equation and make sure it equals 2400.

I got u till 2400=xy
P= 2X+2Y
2Y=P-2X
Y=(P-2X)/2
2400= x (P-2X)/2
but then I'm stuck

2*L^2 + 4*L*H = 2400 cm^2 is the area of all the sides

L^2 Represents the Area of each end thus times two

4 Rectangular sides at L*H

Add them all together you get 2400

We rearrange them so that it ends with zero so it looks kind of like a quadratic equation ax^2 + bx + c = 0.

We could solve it for H at this point, but that would be useless because we don't need every possible value of H. So solve the volume equation for H and put it every where there is an H.

Simplify

In this case we need to solve for volume if we want to find its maximum. So we move the whole term with v in it to the 0 side via subtraction, then we divide out the rest that we don't want.

(-2/4)L^3 + 600L = V

that's about what it should look like once its simplified. There are two x intercepts (V = 0). Somewhere between them is your maximum.

The graph shows it going positive in one direction and negative in another. These parts of the graph don't matter because on the left you can't have a negative Length and on the right you can't have a negative volume.

If you Like you can IM me for more steady answers...

Do I need to start all over again in here? I can if any of you want, OK? Maybe the way I do it is easier... Plus, InfoJunkie,4Life, you can make use of the LaTeX to make your equations easier to understand. It's really simple. The [math] tag are used. You have a tutorial in the sticky of the Math and Sciences forum. :)

Yes please, if u could explain it in simpler terms. That would be great!

Sorry, I didn't even know bbcode supported that.

Here we go.

We know that the area is equal to 2400. However the equation for that amount is still very necessary. To find area of one side we like to use LW = A, where L = Length and W = Width. In order to find the area of a Rectangular box we need to add up the area of each side.

For this we are going to add another variable in order to show where sides are the same measure, and because it is three dimensional we will use H for height.

http://math.asu.edu/~checkman/F2003/113/box.jpg

If you look at the image you will notice that each of box's faces have one opposite of them that is exactly the same size. Normally that would give us the following equation for the area:

2LW + 2HW + 2LH = 2400

Even though the left goes into infinity, it doesn't matter because X Width cannot be negative and even though it goes to negative infinity to the right it does not matter because Y Volume cannot be negative either.

The maximum is going to be between the two x intercepts. V = 0.


The fact that the ends are square helps us out a lot. That means that L = W. This means we can substitute L for W, which would result in the following equation:

2LL + 2HL + 2LH = 2400

Simplified:

2L^2 + 4LW = 2400

Our next important equation is that for volume LWH = V where L, W, and H are the same and V is volume.

Once again H is the same as W and can be supplemented in, resulting in:

LLW = V

Simplified:

L^2W = V

Using the equation for area we could never find maximum volume because it doesn't even have volume in it. The volume equation we just made couldn't find it because it is dependent on the unknown L and W, meaning there are infinite possibilities.

So to achieve the question, we limit the volume equation by the area equation. This is kind of like saying the volume equation depends on the results of the area equation. This is done by putting the volume equation inside of the area equation. I'll show you how.

Solve L^2W = V for W:

\frac{L^2W}{L^2} = \frac{V}{L^2}

Remember: What ever you do to one side of the equation you must do to the other.

Simplified:

W = \frac{V}{L^2}

Now we have an equation for W that we can substitute into the area equation.

Hint: By doing this, we have placed the volume equation inside of the area equation.

2L^2 + 4L\frac{V}{L^2} = 2400

Simplified1:

2L^2 + \frac{4LV}{L^2} = 2400

Simplified2:

2L^2 + \frac{4V}{L} = 2400

The next thing to do is find volume:

First we need to have volume on one side of the equation in order to solve for it.

2L^2 + \frac{4V}{L} - 2400 = 2400 - 2400

2L^2 + \frac{4V}{L} - 2400 = 0

2L^2 + \frac{4V}{L} - \frac{4V}{L} - 2400 = 0 - \frac{4V}{L}

2L^2 -2400 = -\frac{4V}{L}

2L^2 - 2400 = -V\frac{4}{L}

2L^2 - 2400 = -V\frac{4}{L}

\frac{L}{-4}(2L^2 - 2400) = V\frac{-4}{L}\frac{L}{-4}

Hint: A number times its inverse is always 1

\frac{L}{-4}(2L^2 - 2400) = V1

\frac{L(2L^2 - 2400)}{-4} = V

\frac{2L^3 - L2400}{-4} = V

Each part divided by 4:

-\frac{2}{4}L^3 - 600L = V

If we graphed that it would look something like:

http://www.pocketmath.net/articles_imgs/163/algebr19.jpg

Note: The numbers are not correct.

Even though the Left of the equation goes to infinity, it doesn't matter because width cannot be negative. And the left goes to negative infinity, so likewise, volume cannot be negative. The maximum must lie between the two x intercepts; which is where V = 0. You can use Limits, Derivatives, or Tables to find the maximum after this.

If you want to know more just let me know.

Any specific questions on the previous post, just let me know.

Its not as tedious as I wrote it to be. Most of these can be done in your head with some practice. I just used detailed steps to help you follow what I was doing.

okay! Thanks it's much clearer now! Thank you ^_^

No problem...

Good! I see you did it well InfoJunkie4Life! :)

And, you got your answerChristal24 :)