Optimization word problem involving perimeter and area of an arched window

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I'm working through an optimization problem in my textbook about maximizing the area of a rectangular window with an arched top.

My question is concerns how to think about a certian variable. I'm torn because my textbook offers a point of view I disagree with. This almost always means I'm wrong of course, but I can't figure out why.

problem details:

So picture a box under a semicircle.

Starting with area and perimeter, I get $A=\frac12 \pi r^2 +2rh$, and $p= 2r+2h+\pi r$. Solving for h in p I get $h=\frac12 \left[ p-(2+ \pi )r \right]$, leading to $A(r) = pr-(2+ \frac12 \pi ) r^2$ ... Speeding through this because none of it is controversial.

My question is about the fate of p when differentiating A wrt r. Specifically whether p should be considered a constant or not.

The book says constant, so when differentiating A(r) it gives a tidy answer wrt p.

My feeling is p would be more accurately differentiated with respect to r when A'(r) is calculated.

the derivative of A(r) as the book has it

$$ \begin{align} A(r) &= pr-(2+ \frac12 \pi ) r^2 \longrightarrow\\ A'(r) &= p - 2(2 + \frac12 \pi)r \\ &= p - (4 + \pi)r \\ \end{align} $$

the derivative of A(r) as I have been looking at it

$$ \begin{align} A(r) &= pr-(2+ \frac12 \pi ) r^2 \longrightarrow \\ \tfrac{d}{dr} A(r) &= \tfrac{d}{dr} pr - \tfrac{d}{dr} (2 + \frac12 \pi)r^2\\ \tfrac{dA}{dr} &= \tfrac{dp}{dr} r + p \cdot 1 - (4+\pi)r \\ \tfrac{dA}{dr} &= \tfrac{dp}{dr} r + p - (4+\pi)r \\ \end{align} $$

When I consider the perimeter I think of it as non-constant. If we were asked to differentiate p we would have done so with respect to r and h.

This all strongly suggests to me that the books answer to A(r) is wrong... but as I say usually the yolk’s on me!