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Problem: A child wants to build a tunnel using three equal boards, each 4 feet wide, one for the top and one for each side. The two sides are to be tilted at equal angles θ to the floor. What is the maximum cross-sectional area A that can be achieved?

So I drew the triangle and separated it into three parts 2 identical triangles and one rectangle. The areas are then

Rec/Squ. : 4y Triangles: ((1/2)xy)*2

I added them to get the area of the trapezoid. I got (4+x)y

I then wanted to have one variable so that when I derived to set it equal to zero I would be easier.

To do this I evaluated the angle θ, I saw that sinθ=y/4 and cosθ= x/4 (SOHCAHTOA).

I isolated y and x in their separate formulas. I then put the new values for x and y back into the area for the trapezoid formula I got.

(4+x)*y became (4+4cosθ)(4sinθ) then I derived.

When I got the derivative 12(cosθ+cos^2θ-sin^2θ)

I was way more confused then I hoped to be. I looked up the question online and got different formulas for the area to derive, as well as different substitutions plugged in at different times.

What am I doing wrong?

Also what would the constraints be. If I were to format this into a xy graph would it be correct to assume the constraints be between (0, π/2)?

Herb
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    Try to set things up in terms of cross-section area $A$ in terms of a single variable. Use angle $\theta$ if you like, but either $x$ or $y$ can be used (use Pythagorean formula). Maximum area occurs either at a boundary of the domain or an interior critical point. – hardmath Nov 11 '14 at 03:40

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The formula for the area of a trapezoid is $A = \frac 12 h(b_1+b_2)$ where $b_1, b_2$ are the lengths of the two parallel sides. You are given that the top base $b_1=4$ and are asked to find the bottom base $b_2$. The height comes from the triangles: $h^2+\frac 14(b_2-b_1)^2=4^2$ or $b_2=4+\sqrt {64-4h^2}$. Plug this into the area formula, differentiate with respect to $b_2$, set to zero......

Ross Millikan
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