Geometry/Calculus problem?

Gary

A canvas wind shelter is to be constructed for use on the beach
of an island. It is to have a back, two square
sides, and a top. If 150 square feet of canvas
are available, find the depth of the shelter for
which the space inside is maximized assuming
all the canvas is used.

Answer
Say: depth = P and width = L

surface of left side = right side = P²
surface of top = back = PL
total surface = 2P² + 2PL = 150 â L = (150 - 2P²) / 2P = (75 - P²) / P
volume = V = P²L= P²(75 - P²) / P = 75P - P³
V' = 75 - 3P² = 0
âD = â(0² - 4â(-3)â75) = â900 = 30
P = (0 - 30) / (2â(-3)) = 5
P = (0 + 30) / (2â(-3)) = -5

Negative depth makes no sense, so the only solution is P = 5

errr...calculus help?

oOLaura

A shelter for use at the beach has a back, two sides, and a top made of canvas. Find the dimensions that maximize the volume and require 96 square feet of canvas.

Please show work. Thanks in advance.

Answer
The key to this problem is to realize that the two "sides" of the shelter will be square. This is kind of hard to explain, I worked it out on paper and found that the variables representing the dimensions of the "sides" of the shelter could be swapped around without changing any of the formulae.

From here it is a simple maximization problem.
I'll do most of it for you:

Volume = A x B x C where A, B & C are the height, depth and width of the shelter respectively.
Area = 2 x A x B + A x C + B x C = 96

If A = B as I described above:
Area = 2A^2 + 2AC = 96
Therefore, C = (48 - A^2)/A
Therefore, Volume = A x ( 48 - A^2)

Now, maximize volume with respect to A.

Is this for high school? Regards,
Daniel

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