Altitude of a rectangle

Calculus Level pending

The figure shows a rectangle surmounted by a semi-circle. The diameter of the semi-circle coincides with the upper base of the rectangle. If the perimeter of the figure is 5 π + 20 5\pi +20 , find the altitude of the rectangle if the area of the figure is maximum.


The answer is 5.

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1 solution

Let r r and h h be the radius of the semicircle and altitude of the rectangle, respectively.

P = 5 π + 20 = 2 r + 2 h + π r P=5\pi +20=2r+2h+\pi r

5 π + 20 = 2 r + 2 h + π r 5\pi + 20 = 2r+2h+\pi r

2 h = 5 π + 20 − 2 r − π r 2h=5\pi +20-2r-\pi r

h = 5 π + 20 − 2 r − π r 2 h=\dfrac{5\pi + 20 -2r -\pi r}{2}

A = 2 r h + 1 2 π r 2 = 2 r ( 5 π + 20 − 2 r − π r 2 ) + 1 2 π r 2 = 5 π r + 20 r − 2 r 2 − π r 2 + 1 2 π r 2 A=2rh+\dfrac{1}{2}\pi r^2=2r\left(\dfrac{5\pi +20-2r-\pi r}{2}\right)+\dfrac{1}{2}\pi r^2=5\pi r+20r-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2

d A d r = 5 π + 20 − 4 r − 2 π r + 2 ( 1 2 ) ( π ) ( r ) = 5 π + 20 − 4 r − 2 π r + π r = 5 π + 20 − 4 r − π r \dfrac{dA}{dr}=5\pi + 20 - 4r-2\pi r+2\left(\dfrac{1}{2}\right)(\pi)(r)=5\pi + 20-4r-2\pi r + \pi r=5\pi + 20 - 4r - \pi r

d A d r = 0 \dfrac{dA}{dr}=0

5 π + 20 − 4 r − π r = 0 5\pi+20-4r-\pi r=0

4 r + π r = 5 π + 20 4r+\pi r=5\pi +20

r ( 4 + π ) = 5 π + 20 r(4+\pi)=5\pi +20

r = 5 π + 20 4 + π r=\dfrac{5\pi +20}{4+\pi}

r = 5 r=5

It follows that,

h = 5 π + 20 − 2 ( 5 ) − 5 π 2 = 20 − 10 2 = 10 2 = 5 h=\dfrac{5\pi +20 -2(5)-5\pi}{2}=\dfrac{20-10}{2}=\dfrac{10}{2}=\boxed{\color{#D61F06}5}

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