Question #274597

A church window consisting of a rectangle topped by a semicircle is to have a perimeter p. Find the radius of the semicircle if the area of the window is to be a maximum.


1
Expert's answer
2021-12-03T07:15:16-0500

The rectangle has dimensions: 2r2r and xx.

Tne semicircle: 122πr=πr\frac{1}{2}\cdot 2\cdot \pi \cdot r=\pi \cdot r .

p=2r+2x+πrx=12(pr(π+2))p=2r+2x+\pi \cdot r\\ x=\frac{1}{2}(p-r(\pi+2))

The area of the window is to be a maximum:

S=x2r+12πr2==(pr(π+2))r+12πr2==prr2(π+2)+12πr2S=p2r(π+2)+πr=pπr4rpπr4r=0r(π+4)=pr=pπ+4S=x\cdot 2r+\frac{1}{2}\pi \cdot r^2=\\ =(p-r(\pi+2))\cdot r+\frac{1}{2}\pi \cdot r^2=\\ =pr-r^2(\pi+2)+\frac{1}{2}\pi \cdot r^2\\ S'=p-2r(\pi+2)+\pi r=p-\pi r-4r\\ p-\pi r-4r=0\\ r(\pi+4)=p\\ r=\frac{p}{\pi+4}

If r(,pπ+4),S>0r\in (-\infty, \frac{p}{\pi+4}), S'>0 .

If r(pπ+4,),S<0r\in (\frac{p}{\pi+4},\infty), S'<0 .

S(pπ+4)S(\frac{p}{\pi+4}) - max.

r=pπ+4r=\frac{p}{\pi+4} .



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