Expert: Scotto - 5/7/2008

Question
QUESTION: I want to make a fish tank with a volume of 2 m^3 and I would like the base of
the tank to be a rectangle twice as long as it is wide.  The base and sides of the
tank are to be of glass.  What shape will use the least amount of glass (and so
cost least)?

ANSWER: I have found that the measurements to minimize the cost are
base=13.73657091", length=27.47314182", and height=9.15771394".

However, using a ruler might make this difficult.  I found it easier to make the base 13 3/4", the length 27 1/2", and the height 9 1/8".
This may not be the exact measurements, but it is fairly easy to measure.  This still gives a volume of 2.00 ft^3 (1.997).  The total are is just under 1150 in^2.

This total area, however, could be difficult to find to exact measurements, so you might just go with the base 1 ft., lenght 2 ft., and height 1 ft.  This gives you and area of 1152, which is not much more than trying to minimize the area.

When the cost is looked at, it might be cheaper just to buy glass that is a 2 ft by 4 ft and cutting the pieces out from there.  Once this piece of glass has been bought, it could be cut in half both ways into four pieces.  The fourth piece could be cut in half as well to form both of the ends of the tank.  Get a little of the special glass putty (I don't remember what its called) and it can be put together.  You might want to get some trim of some kind to glue to the top of the fish tank.

Hope this helps you out.



---------- FOLLOW-UP ----------

QUESTION: Sorry this was misleading.  This is a calculus optimization problem our teacher
gave us, I don't really want to build one just know how to figure this problem
out.

Answer
Thank you for being honest and clarifying the problem.

l=length, w=width, h=height, and Volume=v=2=lwh.
The equation for volume v can be solved for height h.

The area is the (front)+(back)+(bottom)+(left side)+(right side).
The area on the front and the back is the same and the area on the left side and right side is the same.  The area on the bottom is just bw.  Do you see what the others are?

Sustitute the value for h that was found from the volume equation (in this way get rid of h from that equation) and the equation is reduced to one of l and w.

Now you can note the the length l is twice the width w, so l=2w can be used on this equation of l and w, thereby reducing it so that w is the only variable.

Once you have gotten this far, take the derivative with respect to w (the only variable left) and set it equal to 0.  Once this has been done, solve for w.  Going backwards through the problem, you can then find the length l and the height h.