# What angle θ maximizes the cross-sectional area of the gutter? A rain gutter is made from sheets of metal 9 in. wide. The gutters have a 3-in base and two 3-in sides, folded up to an angle θ.

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### 2 Answers

Top width of gutter = `3+2*3costheta`

Perpendicular distance between to and bottom = `3sintheta`

Cross sectional area (A) of the gutter = `(3+(3+2*3costheta))/2*3sintheta`

When the area is maximum then `(dA)/(d(theta)) = 0`

`A = (3+3+2*3costheta)/2*3sintheta`

`A = 9(1+costheta)*sintheta`

`A = 9(sintheta+costheta*sintheta)`

`A = 9/2(2sintheta+2sintheta*costheta)`

`A = 9/2(2sintheta+sin(2theta))`

`(dA)/(d(theta)) = 9/2(2costheta+2cos(2theta))`

When `(dA)/(d(theta)) = 0`

`9/2(2costheta+2cos(2theta)) = 0`

`cos(2theta)+costheta = 0`

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

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

`2costheta(costheta+1)-1(costheta+1) = 0`

`(costheta+1)(2costheta-1) = 0`

`costheta = -1` and `costheta = 1/2`

`costheta = -1`

`theta = pi`

This cannot happen because the angle should be less than 90 to form the gutter.

`costheta = 1/2`

`costheta = cos(pi/3)`

`theta = pi/3`

If area is maximum at `theta = pi/3` then `(d^2A)/(d(theta)^2) <0` at `theta = pi/3`

`(dA)/(d(theta)) = 9/2(2costheta+2cos(2theta))`

`(d^2A)/(d(theta)^2) = 9/2(-2sintheta-4sin(theta))`

`(d^2A)/(d(theta))^2 = -9/2(2sintheta+4sin(theta))`

since sin(pi/3)>0 then `(d^2A)/(d(theta)^2)<0`

*So we have the maximum Area at `theta = pi/3` *

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### User Comments

Assuming θ is the angle from the horizontal to the gutter, the surface area of the gutter is given by S = 3 * 3 * sin θ + 3 * 3 * sin θ * cos θ = 9 * sin θ * (1 + cos θ).

To optimize the surface area we then need to differentiate this expression with respect to θ:

dS/dθ = 9 * cos θ * (1 + cos θ) - 9 * (sin θ)^2

and find the stationary points of S by solving for 0:

0 = 9 * cos θ * (1 + cos θ) - 9 * (sin θ)^2

=> 0 = (cos θ )^2 - (sin θ)^2 + cos θ

This is satisfied at θ = 60 degrees (with a surface area of ~11.7 square inches)