Area of a rectangle=lw

Area of a semicircle=πr^{2}/2

r=w/2

A_{s}=π(w/2)^{2}/2=πw^{2}/8

Perimeter of the window=2l+w+1/2C

C=2πr=2πw/2=πw

P=2l+w+πw/2

49=2l+w+πw/2

l=(49-w-πw/2)/2

Total Area:

A=lw+πw^{2}/8

sub in for l so whole area is in terms of w

A=((49-w-πw/2)/2)(w)+πw^{2}/8

A=49/2w-w^{2}/2-πw^{2}/4+πw^{2}/8

A=49/2w-w^{2}/2-πw^{2}/8

Find dA/dw

dA/dw=49/2-w-πw/4

set equal to 0

0=49/2-w-πw/4

w(1+π/4)=49/2

w=49/(2+π/2)≈13.7

Do sign test to confirm max:

f'(10)=49/2-10-10π/4≈6.6

f'(15)=49/2-15-15π/4≈-2.3

Since f'(x) is positive to the left of 49/(2+π/2) and negative to the right x=49/(2+π/2) is a maximum.