Let 𝑟 be a radius of a semicircle, H be a distance from top to bottom. Then the horizontal side of the rectangle is equal 2𝑟, vertical side of the rectangle is equal 𝐻 − 𝑟. Effective area of the window is
S=Srect+kSsemi=2r(H−r)+k⋅21πr2 Given
H=2 m,k=21,0<2r≤1.5 m Substitute
S=S(r)=2r(2−r)+21⋅21πr2
S=S(r)=4r−2r2+41πr2,0<r≤0.75 Find the critical number(s)
S′=(4r−2r2+41πr2)′=4−4r+21πr
S′=0=>4−4r+21πr=0r=8−π8 First derivative test
If 0<r<8−π8,S′>0,S increases.
If r>8−π8,S′<0,S decreases.
The function S(r) has a local maximum at r=8−π8.
Since the function S has the only extremum, then the function S has the absolute maximum r=8−π8.
8−π8>1.5 Hence we have to take r=0.75m
The width of the rectangular portion of the window is 2r=1.5m.
The height of the rectangular portion of the window is 2−0.75m=1.25m
The dimensions of the rectangular portion of the window that lets through the most light are
1.5m×1.25m
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