I've seen plenty of norman window threads online and on this platform as well, though I'm stumped on a slightly different version of the usual question. Generally a perimeter is provided and we are asked to optimise the area. I understand the solutions presented to this type of question but I've been given the following problem and have been so far unable to solve it.
Determine the smallest possible perimeter if the area of the window is fixed at 6.8 square metres.
I've developed a formula for both area and perimeter and have isolated the variable 'h': h=(13.6-πr2)/2r. After substituting that into 2h+r+πh, which was my formula for perimeter, I am left with f(r)=(68+5r2)/5r. I took the derivative and got f'(r)=(5r2-68)/5r2. But this still doesn't help me out, as the function never reaches zero. I'm totally bewildered. I'm not sure if I've gone about this the right way so I'll leave a drawing of the problem and the variables and my working here. Please excuse the terrible drawing. Thanks
Update: I accidentally treated r as the diameter in some of my calculations but the end result remains the same, the formula for perimeter cannot be proven correctly with the results obtained and the derivative's domain cannot equal 0.
After googling "Norman window", it appears to be a rectangle surmounted by a semicircle.
Hint
With $r$ being the radius of the semicircle and $h$ being the height of the rectangle, we can write the perimeter $P$ and area $A$ as $$P = \pi r + 2h + 2r$$ $$A= \frac{\pi r^2}{2} + 2rh = 6.8$$ From the equation for area we find that $$h = \frac{6.8}{2r} - \frac{\pi r}{4}$$ Inserting this into the perimeter equation we find $$P=r(\frac{\pi}{2}+2)+\frac{6.8}{r}$$ Now differentiate this equation, set it to zero, find $r$ and you are done.