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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2022 · Mathematics · Application of Derivatives

A running track of 440 ft is to be laid out enclosing a football...

A running track of 440ft440 \mathrm{ft} is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then the lengths of its side are

Options

  1. A70ft70 \mathrm{ft} and 110ft110 \mathrm{ft}
  2. B80ft80 \mathrm{ft} and 120ft120 \mathrm{ft}
  3. C35ft35 \mathrm{ft} and 110ft110 \mathrm{ft}
  4. D35ft35 \mathrm{ft} and 120ft120 \mathrm{ft}

Correct answer

C. 35ft35 \mathrm{ft} and 110ft110 \mathrm{ft}

Step-by-step solution

Perimeter =440ft⇒2x+πr+πr=4402x+2πr=440=440 \mathrm{ft}\Rightarrow \quad 2 x+\pi r+\pi r=4402 x+2 \pi r=440A= Area of  the rectangular portion =x2rA=x(440−2x)π=1π(440x−2x2)\begin{aligned} A & =\text{ Area\text{ of } the rectangular portion }= x 2 r \\ A & =x \frac{(440-2 x)}{\pi}=\frac{1}{\pi}\left(440 x-2 x²\right) \end{aligned}LetdAdx=1π(440−4x)=0Let \frac{d A}{d x}=\frac{1}{\pi}(440-4 x)=0⇒x=110 for which d2Adx2<0⇒A is maximum when x=110⇒2r=440−2xπ=440−22022/7=70 and ∴r=35ft and x=110ft\begin{aligned}& \Rightarrow x=110 \text{ for which } \frac{d^2 A}{d x^2}< 0 \\ & \Rightarrow A \text{ is maximum when } x=110 \\ & \Rightarrow 2 r=\frac{440-2 x}{\pi}=\frac{440-220}{22 / 7}=70 \\ \text{ and } \therefore \quad r=35 \mathrm{ft} \text{ and } x=110 \mathrm{ft} \end{aligned}$$
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