A rain gutter is to be constructed from a metal sheet of width 3w by bending up one third of the sheet on each side through an angle theta. How should theta be chosen so that the gutter will carry...

A rain gutter is to be constructed from a metal sheet of width 3w by bending up one third of the sheet on each side through an angle theta. How should theta be chosen so that the gutter will carry the maximum amount of water?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The rain gutter is to be created with a sheet with a width 3w by bending up to one third of the sheet on each side through an angle `theta` .

To determine the optimum angle `theta` that maximizes the volume of water, the problem can be considered to be one where a sheet of length 3w has to be bent on either side by an angle `theta` and the area of the resulting open figure is maximum.

When the sheet is bent the resulting figure consists of a rectangle between two triangles. The dimension of the rectangle is w*w*sin `theta` . The area of each triangle is (1/2)*w*sin (`90 - theta`) *w*cos(90 -` theta`) .

Adding the three the total area is A = w^2*sin `theta` + w^2*sin(90- `theta` )*cos (90-`theta` )

=> A = w^2*sin `theta` + w^2*cos `theta` *sin `theta`

To maximize area solve `(dA)/(d theta ) = w^2(sin theta + -sin^2 theta + cos^2 theta )` = 0

`w^2(cos theta + cos^2 theta - 1 + cos^2 theta ) = 0`

=> `2*cos^2 theta + cos theta - 1 = 0`

=> `2*cos^2 theta + 2*cos theta - cos theta - 1 = 0`

=> `2*cos theta(cos theta + 1) - 1(cos theta + 1) = 0`

=> `cos theta = 1/2` and `cos theta = -1`

=> `theta` = 60 degrees and `theta` = 180 degrees

The angle cannot be 180 degrees.

The required angle `theta` by which the sheet should be bent is equal to 60 degrees.

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