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September 16, 2014

September 16, 2014

Posted by **soffy** on Monday, June 11, 2012 at 10:49am.

- calculus -
**Steve**, Monday, June 11, 2012 at 11:06amFor lack of specificity, I will let θ be the angle between the upper base and a side. I expect it will end up 90°, making a square cross-section, but let's wee what happens.

The height of the trapezoid will be 10sinθ, so the area is

a = 10sinθ (10+10+2*10cosθ)

= 10sinθ (20+20cosθ)

= 200sinθ (1+cosθ)

da/dθ = 200cosθ (1+cosθ) + 200sinθ(-sinθ)

= 200(cos^2θ + cosθ - sin^2θ)

= 200(2cos^2θ+cosθ-1)

da/dθ = 0 when cosθ = 1/2 or -1.

Looks like the max area occurs at θ=π/3

Looks like my gut feeling was wrong. Good thing, as it would have been a boring trapezoid.

- calculus - correction -
**Steve**, Monday, June 11, 2012 at 11:07amOops. I forgot to divide by 2 when getting the average of the bases, but that does not affect the answer in this case.

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