# trapeze-shaped profile: what values do a, h and alpha have to have so that the moistened perimeter of the canal is minimal

(english is not my first language, so sorry if this text isn't fluently readable) We have a trapezoid with area $$A$$ given: see the picture. The question is: How to choose width $$a$$, height $$h$$ and angle of slope $$\alpha$$ so that the moistened perimeter is a minimum? (b is needed to express the minimum of moistened perimeter as 2b + a. )

My thought was to create a function $$f: \Bbb R^3 \to\Bbb R$$ with $$a$$,$$h$$ and $$\alpha$$ being its parameters. But I have really no clue how $$f$$ could look like. I already tried something like find a way to express $$A$$ with the three given Variables $$a,h$$ and $$\alpha$$ and then make $$f$$ express the moistened perimeter $$(2b +a)$$ divided by $$A$$ so that if we look at the extreme values of $$f$$ we get that a minimum of that function might be a possible solution the the problem as a whole. But the function I got looks very ugly and i really don't know if that's what's needed...
My $$f$$ looks like this: $$f(a,h,\alpha) = \frac{2b + a}{A} = \frac{\frac{2h}{\sin\alpha} + a}{ A }$$ by using the fact that $$b = \frac{h}{\sin\alpha}$$, and $$f = \frac{\frac{2h}{\sin\alpha} + a } {ah + \frac{h^2\sin(90-\alpha)}{\sin\alpha}}$$ by using the fact that $$A = (a+c)\cdot \frac{h}{2}$$ and $$c = a + 2 \cdot \frac{h\cdot\sin(90-\alpha)}{\sin\alpha}$$.

Guide:

From your text, it seems that you want to minimize $$2b+a$$ subject to

$$A=ah+2b\cos \alpha \sin\alpha$$

Here $$A$$ is given, and we get to decide $$a,h, \alpha$$.

$$2b=\frac{A-ah}{\sin\alpha\cos\alpha}$$

Hence, we want to minimize

$$\frac{2(A-ah)}{\sin 2\alpha}+a=2(A-ah)cosec 2\alpha + a$$

Differentiate with respect to $$a$$,

$$-2h cosec2 \alpha +1 = 0$$

Differentiate with respect to $$h$$,

$$-2acosec 2 \alpha =0$$

Differentiate with respect to $$\alpha$$,

$$-2(A-ah)\cot \alpha cosec \alpha=0$$

Try to solve for $$a, h, \alpha$$ from the equations.