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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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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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)
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