I need help. I don't know how to solve this Maxima-minima problem using derivatives:?

Question

A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner?

Answer 1

The length of the pipe is given by this equation:

L = 9sec(x) + 6csc(x)

where x is the angle made by the pipe and the walls of the 6' hallway and the complement of the angle made by the pipe and the walls of the 9' hallway. Taking the first derivative:

dL/dx = 9sec(x)tan(x) - 6csc(x)cot(x)
= 9sec(x)tan(x) - 6/[sec(x)tan(x)]

Set that equal to 0 and solve for x. Then substitue x into the original equation for L.

0 = 9sec^2(x)tan^2(x) - 6
2/3 = sec^2(x)tan^2(x)
2/3 = tan^2(x)/cos^2(x)
2/3 = [1 - cos(2x)]/[1 + cos(2x)] * 2/[1 + cos(2x)]
1/3 = [1 - cos(2x)]/[1 + cos(2x)]^2
[1 + cos(2x)]^2 = 3[1 - cos(2x)]
1 + 2cos(2x) + cos^2(2x) = 3 - 3cos(2x)
cos^2(2x) + 5cos(2x) - 2 = 0

Let cos(2x) = u. Then

u = [-5 +- sqrt(33)]/2

Only the positive answer makes sense, so

cos(2x) = 0.372281323
x = 0.594665255 radians, around 34 degrees.

Substituting back into the equation for L gives

L = 21.57501337 feet

Final answer.

Answer 2

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