$A = 10m^2$
Shape: semicircle on top of rectangle Let $x$ = radius of semicircle
Therefore, length of rectangle's base = $2x$
Let $h_{rectangle} = y$
Therefore perimeter $(P) = 2x + 2y + πx$
$A_{semicircle} = \frac{πx^2}{2}$
$A_{rectangle} = 2x·y$
At this point I'm not entirely sure where to go. I know I've set up the perimeter equation correctly and that I need to differentiate it, most likely with respect to $x$. But do I have to make a substitution first?
Start by identifying the constraint and what you're trying to optimize/maximize:
Constraint: $A = 2xy+\frac{\pi x^2}{2} = 10$
Maximize: $P = 2y+2x+\pi x$
From here, solve the constraint for $y$ so that we can make the maximization equation a function of $x$: $$10 = 2xy+\frac{\pi x^2}{2} \\ 2xy = 10-\frac{\pi x^2}{2} \\ y = \frac{5}{x}-\frac{\pi x}{4}$$
Plug this into the maximization equation: $$2\left(\frac{5}{x}-\frac{\pi x}{4}\right)+2x+\pi x \longrightarrow \frac{1}{2} (4+\pi ) x+\frac{10}{x}$$
Then take the derivative of this equation: $$\frac{d}{dx}P=\frac{4+\pi }{2}-\frac{10}{x^2}$$
Set it equal to $0$ to find critical points: $$x = \pm 2 \sqrt{\frac{5}{4+\pi}}$$
Plug it back into the equation for the perimeter: $$P\left(2 \sqrt{\frac{5}{4+\pi}}\right) = 2 \sqrt{5 (4+\pi )} \approx 11.9512$$