Skip to content
Question

Q.Case Study – 2 Read the following passage and answer the questions given below : "In an elliptical sports field, the authority wants to design a rectangular soccer field with the maximum possible area. The sports field is given by the graph of x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1."

(i) If the length and breadth of the rectangular soccer field be 2x2x and 2y2y respectively, then find the area function A(x)A(x) in terms of xx. [1]
(ii) Find the critical point(s) of the area function A(x)A(x). [1]
(iii)
(a) Using first derivative test, find the length 2x2x and breadth 2y2y of the soccer field (in terms of aa and bb) that maximize the area. [2]
(OR)
(b) Using second derivative test, find the length 2x2x and breadth 2y2y of the soccer field (in terms of aa and bb) that maximize the area. [2]
CBSECBSE Class XII Board 2024Subjective· 4mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Area A(x)=4baxa2−x2A(x)=\dfrac{4b}{a}x\sqrt{a^2-x^2}; setting A′(x)=0A'(x)=0 gives x=a2x=\dfrac{a}{\sqrt2}, and both derivative tests confirm a maximum, giving length 2 a\sqrt2\,a and breadth 2 b\sqrt2\,b.

For the rectangle of sides 2x,2y2x,2y inscribed in x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1: y=baa2−x2y=\dfrac{b}{a}\sqrt{a^2-x^2}; maximise A=4xyA=4xy using A′(x)=0A'(x)=0 and the first/second derivative test.

(i) Area function

  1. Area =(2x)(2y)=4xy.=(2x)(2y)=4xy. From the ellipse, y=baa2−x2.y=\dfrac{b}{a}\sqrt{a^2-x^2}.
  2. A(x)=4x⋅baa2−x2=4ba xa2−x2.A(x)=4x\cdot\dfrac{b}{a}\sqrt{a^2-x^2}=\dfrac{4b}{a}\,x\sqrt{a^2-x^2}.

(ii) Critical point(s)

3. A′(x)=4ba[a2−x2+x⋅−xa2−x2]=4ba⋅a2−2x2a2−x2.A'(x)=\dfrac{4b}{a}\left[\sqrt{a^2-x^2}+x\cdot\dfrac{-x}{\sqrt{a^2-x^2}}\right]=\dfrac{4b}{a}\cdot\dfrac{a^2-2x^2}{\sqrt{a^2-x^2}}.

4. A′(x)=0⇒a2−2x2=0⇒x2=a22⇒x=a2A'(x)=0\Rightarrow a^2-2x^2=0\Rightarrow x^2=\dfrac{a^2}{2}\Rightarrow x=\dfrac{a}{\sqrt2} (taking x>0x>0). Critical point: x=a2.x=\dfrac{a}{\sqrt2}.

(iii)(a) First derivative test

5. For xx slightly less than a2\dfrac{a}{\sqrt2}, a2−2x2>0a^2-2x^2>0 so A′(x)>0A'(x)>0; for xx slightly greater, a2−2x2<0a^2-2x^2<0 so A′(x)<0A'(x)<0. Sign change +→−+\to- means AA is maximum at x=a2.x=\dfrac{a}{\sqrt2}.

6. Then y=baa2−a22=ba⋅a2=b2.y=\dfrac{b}{a}\sqrt{a^2-\dfrac{a^2}{2}}=\dfrac{b}{a}\cdot\dfrac{a}{\sqrt2}=\dfrac{b}{\sqrt2}. Length 2x=2a2=2 a2x=\dfrac{2a}{\sqrt2}=\sqrt2\,a; breadth 2y=2b2=2 b.2y=\dfrac{2b}{\sqrt2}=\sqrt2\,b. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.