A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum amount of water?

Question

Physics

Elisha Frost

27 June, 02:19

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum amount of water?

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Answers (1)

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Sheldon Mcclain · Answer 1

60

Explanation:

The volume of the water carried will be proportional to the cross-sectional area. The cross-section is a trapezoid with height 10sint, where t represents theta. The top of the trapezoid is 10 + (2) (10cost), i. e. 10 + 20cost. The base of the trapezoid is 10.

Area of trapezoid = (average of bases) times (height)

= ([10 + (10 + 20cost) ] / 2) 10 sint

= (10 + 10cost) (10sint)

= 100sint + 100sintcost, call this A (t). You need to maximize. So differentiate and set equal to zero.

dA/dt = - 100sin (t) ^2+100cos (t) + 100cos (t) ^2 = 0, divide by 100:

-sin (t) ^2 + cos (t) + cos (t) ^2 = 0, replace sin^2 by 1-cos^2

2cos (t) ^2 + cos (t) - 1 = 0, factor

(1+cos (t)) (2cos (t) - 1) = 0, so

cos (t) = 1, t=0 that give a min (zero area) not a max, or

cos (t) = 1/2, so t=60 degrees. This gives the max.