An optimization problem: If 1200 centimeters squared of material is
available to make a box with a square base and an open top, find the
largest possible volume of the box.

Here's how to do the problem:

Let the length, width, and height of the box (in cm) be represented by
x, x, and y (because the base is a square, the length and width must
be equal):

V = x^2*y  (volume formula)

We can express V as a function of just one variable x by using the
fact that the area of the four sides and the bottom of the box is:

4xy + x^2 = 1200

Solving for y, we get:

(1200 - x^2)/4x = y

Now plug in to the volume formula:

V = x^2*y = x^2*(1200 - x^2)/4x = 300x - (x^3)/4

Now comes the main part of the problem. As you may or may not recall,
finding maximum/minimum values ("extremes") requires the use of
partial derivatives (because we are only working with one variable,
this is more of a regular derivative):

dV/dx = 300x - (x^3)/4
    0 = 300x - (x^3)/4
    0 = x(300 - (x^3)/4)

    It's 0 only when x = 0 or x = 10.62

Because a side length can't be zero, x (length & width) must be 10.62cm.

So, the box will have a maximum volume when both the length and width
are 10.62 cm. The height (y) can be found using the formula derived
above:

y = (1200 - x^2)/4x = 25.59cm

Now that we know the dimensions, we can find the volume:

V = x^2 * y = 2886.2 cm^3


Hope this helps! (and you didn't have to pay me $2.50)

The largest volume box will be a cube, i.e., bottom and four sides of
equal size and square.
1200/5 = 240   Square root of 240 = 15.49, the length of the sides, ^3
= 3718 cubic cm.
Of course, you will have to cut and paste a bit to use all the material.

How does one know that a cube is the optimal shape?
A sphere is the figure with the best volume to surface ratio, so logic
(common sense) indicates that the rectilinear figure closest to a
sphere in proportions will have the largest volume to surface ratio.
Or can you prove me wrong?

jonsha- Thanks, your answer helped me out of my mental block! Even
though when you took the derivative of the voume equation, it should
be dV/dt = 300 - 3/4 x^2 and not what you put, dV/dx = 300x - (x^3)/4
But, I was able to figure out myself where it went wrong.
As for myoarin-I'm not sure you ever have done one of these problems.
At least not for a Calculus class. Maybe your common sense told you
"that the rectilinear figure closest to a
sphere in proportions will have the largest volume to surface ratio,"
but my common sense told me to use a box because the words in the
problem clearly ask to solve the problem using a figure that is a box.

Here is the answer.

Jonshua started out correctly.

Let the length, width, and height of the box (in cm) be represented by
x, x, and y (because the base is a square, the length and width must
be equal):

V = x^2*y  (volume formula)

We can express V as a function of just one variable x by using the
fact that the area of the four sides and the bottom of the box is:

4xy + x^2 = 1200

Solving for y, we get:

(1200 - x^2)/4x = y

Now plug in to the volume formula:

V = x^2*y = x^2*(1200 - x^2)/4x = 300x - (x^3)/4

Now comes the main part of the problem. As you may or may not recall,
finding maximum/minimum values ("extremes") requires the use of
partial derivatives (because we are only working with one variable,
this is more of a regular derivative):

Here is where a correction needs to be made

V = 300x - (x^3) / 4

dV/dx = 300 - 3/4 *x^2 = 0
    3/4 *x^2 = 300
         x^2 = (4/3)* 300 = 400
           x = 20

y = (1200 - x^2)/4x = (1200 - (20^2) / (4 * 20)
  = (1200 - 400) / 80 = 800 / 80 = 10

V = x^2*y = (20^2) * 10 = 400 * 10 = 4000

As a check on the answer

Surface area of the box

4xy + x^2 = (4*20*10) + (20^2) = 800 + 400 = 1200

Hope this helps.

OH well, so much for my kind of common sense!  :-/

Myoarin,

For a closed box, it would have been in the shape of a cube, but there
was no top.  Think of it this way - you want to maximize the volume
you get for a given surface area.  You have walls and bases (top and
bottom).  You have to enclose all the walls (y variable) but only one
of the bases (x variable), namely the bottom.  Consequently you get
more bang for your buck (volume for your surface area) when you
increase the base than you do when you make the walls taller.

HI, Anse1001,
Thank you for explaining the error in my thinking.  Of course ...!  
Myoarin