# calculus max area

Maximize volume of a pyramid cut from peice of paper.

(Problem) A pyramid consists of 4 isosceles triangles around a square base. If this is to be cut and folded out of a single square piece of paper, Maximize the volume.

I am not sure where to begin here. I know I am going to have to use the volume of the pyramid but I don't know what else to do.

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1. First take a piece of paper and draw the developed surface of the pyramid, namely a square base, with 4 isosceles triangles folded flat on each side.

You will want the vertices of the four triangles (equivalent to the top of the pyramid) form a square, which is your paper.

Draw a diagram of a piece of paper with side L. Draw the unfolded pyramid inside the square, with vertices placed at the corners of the square.

Label the side of the base as x, and the slant height of the pyramid as l.

See, for example:
http://img525.imageshack.us/img525/2720/1343935243.jpg

The diagonal of the square paper is therefore √2*L.

Add up across the diagonal gives
x+2*l=√*L which gives
l=(√*L-x)/2

The vertical height h of the pyramid is
given by Pythagoras theorem as
h²=l²-(x/2)²
or
h=√(l²-(x/2)²)

The volume V of the pyramid is
V=(1/3)x²h
=(1/3)x²√(l²-(x/2)²)

Differentiate V with respect to x and equate to zero to get the maximum/minimum volume.

Among the solutions, you will find x=0 gives a volume of zero (minimum), and the other value of x=sqrt(8)L/5 gives the maximum volume, which should come up to:
V=(8*L³)/(75*sqrt(10))

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