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1. There is a wall that does not need fencing [at x=0]. The curve that also does not need fencing is y=x^2/9. The fence is four units long and extends from the y axis horizontally some distance and then drops down to the curve. Find the lengths of the horizontal and vertical parts of the fence to maximize the area. [Let the height that the fence starts on the y axis be designated as h. Now find the area in terms of h and other variables. Then get rid of h.

This is a major Calculus question, requiring both integration and differentiation, but it will work out pretty neat.

I am going to the start of the horizontal P(0,h), the right end of the horizontal Q(a,h) and the point where the vertical meets R(a,(a^2)/9)

then PQ + QR = 4

a + (h-(a^2)/9 = 4

h = (a^2 - 9a + 36)/9

The area of the region between the curve y = x^2 /9 and the x-axis from 0 to a is
⌠ x^2 /9 dx from 0 to a, which is (a^3)/27

So the area of your question would be

A = ah - (a^3)/27
=a(a^2 - 9a + 36)/9 - (a^3)/27
= 1/9(a^3 - 9a^2 + 36a) - (a^3)/27

I then found dA/da and got

1/9(3a^2 - 18a + 36) - (1/9)a^2

setting this equal to zero for a max/min of area I got the nice quadratic

a^2 - 9a - 18 = 0
(a-6)(a-3)=0

so a = 6 or a=3, but the total of the two straight line segments has to be 4, so a=6 is extraneous

then a=3, which makes h=2 and the area=5