derivative...max value?

Question

There is a light source located over the center of a circular table of diameter 4 ft. find the heigh h of the light source such that the illiumination I at the perimeter of the table is maxium if I=k(sina)/s^2 where s is the slant height, a is the angle at which the light strikes the table, and k is a constant.
so what i did was take the I equation and solve for h and s using the sin of a whcih is h/s in the triangle, and solved for s, and i solved for h using the tan of a which is h/2...i got the equation
k sin a
--------
4sec^2(x)
now i dont know what to do.
im supposed to take the derivative, but its messy with the sina over sec squared x.
what do i do/how do i take the derivative?

correction, it should be sec^2 a not x.

Answer 1

The first thing I would do is rewrite both a and s in terms of h.

Then you use the following rules to find the derivative.

1. Use the quotient rule to find the derivative of sin(a)/s^2. If you don't know the quotient rule -- look it up and memorize it.

2. The derivative of the sin is the cosin -- use the chain rule to get cos(a)*da/dh

3. The derivative of the bottom is 2*s * ds/dh (again using the chain rule).

Once you get the derivative, solve for zero.

Good luck.