(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 $c = a + 2 \cdot \frac{h\cdot\sin(90-\alpha)}{\sin\alpha}$.

I'm happy about basically any advice/thoughts.



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 $a,h, \alpha$.


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.


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