A running track 440 ft, is to be laid out enclosing football field the shape of which a rectangle with a semicircle at each end. If the area of the rectangular position is to be maximum then the dimensions of the rectangle are

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110, 60

100, 80

100, 70

110, 70

answer is B.

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Detailed Solution

$\begin{array}{l} 2 x + 2 π y = 440 \Rightarrow x + π y = 220 \\ \Rightarrow A r e a = x (2 y) = (220 - π y) (2 y) \Rightarrow f (y) = 440 y - 2 π y^{2} \\ f^{1} (y) = 0 \Rightarrow 440 = 4 π y \Rightarrow 110 = \frac{22}{7} y \Rightarrow y = 35 \end{array}$

Therefore, length of the rectangle is $x = 110$ and width of the rectangle is $2 y = 2 (35) = 70$

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