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 lengthof the longest pipe that can be carried horizontally around the corner.

Answer 1

I hope x, y and the angle θ are clear.

The hypothenuse of the upper triangle is 6/sinθ or 6 cscθ. Similarly the hypothenuse of the lower triangle is 9/cosθ or 9 secθ. Therefore, the lengthof the bar is 6 cscθ +9secθ. Differentiating this with respect to θ we get

-6cscθ + 9secθ

Setting this equal to zero and changing the trigonometric identities

(9/cosθ).(sinθ/cosθ) - (6/sinθ).(cosθ/sinθ) = 0

(9sinθ/cos²θ) - (6cosθ/sin²θ) =0

(9sinθ.sin²θ - 6cosθ.cos²θ) / (sin²θ.cos²θ) = 0

9sin³θ - 6cos³θ = 0

9sin³θ = 6cos³θ

sin³θ/cos³θ = tan³θ = 6/9 =2/3

tanθ = (2/3)^1/3

From here with a scientific calculator θ can be found and from there x and y can be found. The lenght of the bar is the square root of(9+x)² + (6+y)²

I found these values and then the final answer as about 16.6 feet.

My writing may be unclear but if you copy everything on paper with standard form it should be clearer. I am new to this site and I found using theeditor difficult. I hope the answer helps. Best regards...