Arain 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 6. How should theta be chosen so that the gutter will carry the maximum amount of water?

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lalilulelo2k3eq · Accepted Answer

The amount of water that the gutter can carry is proportional to the area of a cross section of the gutter.If h is the height that the tip of one of the bent-up segments rises above the base, then the cross-sectional area isA=10h+2(12bh).Now, by the definition of the sine,sin⁡θ=h10⇒h=10sin⁡θLikewise,cos⁡θ=b10⇒b=10cos⁡θTherefore,A(θ)=10(10sin⁡θ)+(10sin⁡θ)(10cos⁡θ)=100sin⁡θ+100sin⁡θcos⁡θ.This is the function we&#039;re trying to maximize, so differentiate and find critical points:A′(θ)=100(cos⁡θ−(1−cos2θ)+cos2θ)=100(2cos2θ+cos⁡θ−1)=100(2cos⁡θ−1)(cos⁡θ+1).Therefore, A′(θ)=0 when100(2cos⁡θ−1)(cos⁡θ+1)=0of, equivalently, when eithercos⁡θ=12orcos⁡θ=−1.Thus θ=π3orθ=π.Now, theta can range from 0 to π, so we just plug in critical points and endpoints into the area function:A(0)=0A(π3)=100⋅32+100⋅32⋅12=30034≈129.9A(π)=0Therefore, the gutter can carry the maximum amount of water when θ=π3

Raymond Foley · Accepted Answer

Answer: The amount of water that the gutter can carry is proportional to the area of a cross section of the gutter. If h is the height that the tip of one of the bent-up segments rises above the base, then the cross-sectional area is  A=10h+2(12bh)  Now, by the definition of the sine,  sin⁡θ=h10,soh=10sin⁡θ.  Likewise,  cos⁡θ=b10,sob=10cos⁡θ.  Therefore,  A(θ)=10(10sin⁡θ)+(10sin⁡θ)(10cos⁡θ)=100sin⁡θ+100sin⁡θcos⁡θ.  This is the function were

Arain gutter is to be constructed from a metal sheet

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