If a running track of 500 ft. is to be laid | vedclass

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(B) Let the length of the rectangular portion be $x$ and the radius of the semicircular ends be $r$. The total perimeter of the track is given by $P =

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) Let the length of the rectangular portion be $x$ and the radius of the semicircular ends be $r$. The total perimeter of the track is given by $P = 2x + 2\pi r = 500 \ ft$.From this,we get $x + \pi r = 250$,so $x = 250 - \pi r$.The area of the rectangular portion is $A = x 	imes (2r) = (250 - \pi r)(2r) = 500r - 2\pi r^2$.To maximize the area,we find the derivative with respect to $r$ and set it to zero:$\frac{dA}{dr} = 500 - 4\pi r = 0 \Rightarrow r = \frac{125}{\pi}$.Now,substitute $r$ back into the expression for $x$:$x = 250 - \pi \left(\frac{125}{\pi}ight) = 250 - 125 = 125 \ ft$.Thus,the length of the rectangular portion for maximum area is $125 \ ft$.

If a running track of $500 \ ft$. is to be laid out enclosing a playground,the shape of which is a rectangle with a semicircle at each end,then the length of the rectangular portion such that the area of the rectangular portion is to be maximum is (in feet).

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