CBSE Sample Paper 2022-23 · Section E · 4 marks

Q37: Case study: the largest rectangular soccer field in an elliptical sport field

Sample 2022-23Ch 6 · Application of Derivatives

The question

Case-Study 2: Read the following passage and answer the questions given below.

In an elliptical sport field the authority wants to design a rectangular soccer field with the maximum possible area. The sport field is given by the graph of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ⁠.

(i) If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x. (1 mark)

(ii) Find the critical point of the function. (1 mark)

(iii) (a) Use First derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area. (2 marks)

(iii) (b) Use Second Derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area. (2 marks)

Exam answer

(i) Let $(x, y)$ be the upper right vertex of the rectangle, with $x, y > 0$ ⁠. It lies on the ellipse, so $y = \frac{b}{a}\sqrt{a^2 - x^2}$ ⁠.

Why?

Solve the equation of the ellipse for $y$ ⁠:

\begin{aligned} \frac{y^2}{b^2} &= 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2} \\ y^2 &= \frac{b^2}{a^2}(a^2 - x^2) \end{aligned}

The vertex is above the x-axis, so take the positive square root.

\begin{aligned} y &= \sqrt{\frac{b^2}{a^2}}\,\sqrt{a^2 - x^2} \\ &= \frac{b}{a}\sqrt{a^2 - x^2} \end{aligned}

The rectangle of length

2x

and breadth

2y

inside the ellipse, with its vertex

(x, y)

on the ellipse.

The area function is

\begin{aligned} A &= 2x \times 2\frac{b}{a}\sqrt{a^2 - x^2} \\ &= \frac{4b}{a}x\sqrt{a^2 - x^2},\ x \in (0, a) \end{aligned}

Why?

Area of a rectangle = length $\times$ breadth $= 2x \times 2y$ ⁠. Then put in $y$ from above.

\begin{aligned} A &= 2x \times 2y \\ &= 2x \times 2 \cdot \frac{b}{a}\sqrt{a^2 - x^2} \\ &= (2 \times 2)\frac{b}{a}\,x\sqrt{a^2 - x^2} \\ &= \frac{4b}{a}x\sqrt{a^2 - x^2} \end{aligned}

The vertex moves along the ellipse between $x = 0$ and $x = a$ ⁠. At either end the rectangle flattens to zero area, so $x \in (0, a)$ ⁠.

$\therefore\ A(x) = \frac{4b}{a}x\sqrt{a^2 - x^2}$ ⁠, $x \in (0, a)$

(ii)

\begin{aligned} \frac{dA}{dx} &= \frac{4b}{a}\bigg[x \times \frac{-x}{\sqrt{a^2 - x^2}} \\ &\qquad + \sqrt{a^2 - x^2}\bigg] \\ &= \frac{4b}{a} \times \frac{a^2 - 2x^2}{\sqrt{a^2 - x^2}} \\ &= -\frac{4b}{a}\,\frac{2\left(x + \frac{a}{\sqrt2}\right)\left(x - \frac{a}{\sqrt2}\right)}{\sqrt{a^2 - x^2}} \end{aligned}

Why?

Product rule on $x$ times $\sqrt{a^2 - x^2}$ ⁠, with the chain rule for the root:

\begin{aligned} &\frac{d}{dx}\sqrt{a^2 - x^2} \\ &= \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}} \end{aligned}

Write the parts: $u = x$ ⁠, $v = \sqrt{a^2 - x^2}$ ⁠, so $u' = 1$ and $v' = \frac{-x}{\sqrt{a^2 - x^2}}$ ⁠. Then

\begin{aligned} (uv)' &= uv' + vu' \\ &= x \times \frac{-x}{\sqrt{a^2 - x^2}} \\ &\quad + \sqrt{a^2 - x^2} \times 1 \end{aligned}

Over the common denominator $\sqrt{a^2 - x^2}$ ⁠, the numerator is $-x^2 + (a^2 - x^2) = a^2 - 2x^2$ ⁠.

\begin{aligned} &\frac{-x^2}{\sqrt{a^2 - x^2}} + \frac{a^2 - x^2}{\sqrt{a^2 - x^2}} \\ &= \frac{-x^2 + a^2 - x^2}{\sqrt{a^2 - x^2}} \\ &= \frac{a^2 - 2x^2}{\sqrt{a^2 - x^2}} \end{aligned}

Here $\sqrt{a^2 - x^2} = \frac{a^2 - x^2}{\sqrt{a^2 - x^2}}$ ⁠: multiply its top and bottom by $\sqrt{a^2 - x^2}$ ⁠, and $\sqrt{a^2 - x^2} \times \sqrt{a^2 - x^2} = a^2 - x^2$ ⁠.

Factor it as a difference of squares:

\begin{aligned} &a^2 - 2x^2 = -2\left(x^2 - \frac{a^2}{2}\right) \\ &= -2\left(x + \frac{a}{\sqrt2}\right)\left(x - \frac{a}{\sqrt2}\right) \end{aligned}

This uses $\frac{a^2}{2} = \left(\frac{a}{\sqrt2}\right)^2$ and $p^2 - q^2 = (p + q)(p - q)$ ⁠.

\frac{dA}{dx} = 0 \Rightarrow x = \frac{a}{\sqrt2} \quad (x > 0)

Why?

A fraction is zero only when its numerator is zero. The denominator is not zero for $0 < x < a$ ⁠. The other root $x = -\frac{a}{\sqrt2}$ is rejected because $x > 0$ ⁠.

\begin{aligned} a^2 - 2x^2 &= 0 \\ a^2 &= 2x^2 \\ x^2 &= \frac{a^2}{2} \\ x &= \pm\frac{a}{\sqrt2} \end{aligned}

$\therefore\ x = \frac{a}{\sqrt2}$ is the critical point.

(iii) (a) For values of $x$ less than $\frac{a}{\sqrt2}$ and close to $\frac{a}{\sqrt2}$ ⁠, $\frac{dA}{dx} > 0$ ⁠, and for values of $x$ greater than $\frac{a}{\sqrt2}$ and close to $\frac{a}{\sqrt2}$ ⁠, $\frac{dA}{dx} < 0$ ⁠.

Why?

In $\frac{dA}{dx}$ the factors $\frac{4b}{a}$ and $\sqrt{a^2 - x^2}$ are positive, so the sign is the sign of $a^2 - 2x^2$ ⁠.

For $x$ a little less than $\frac{a}{\sqrt2}$ ⁠, $2x^2 < a^2$ ⁠, so it is positive. For $x$ a little more, $2x^2 > a^2$ ⁠, so it is negative.

First derivative test: if $f'$ changes sign from $+$ to $-$ at $c$ ⁠, then $c$ is a point of local maximum.

Hence, by the first derivative test, there is a local maximum at the critical point $x = \frac{a}{\sqrt2}$ ⁠. Since there is only one critical point, the area of the soccer field is maximum at $x = \frac{a}{\sqrt2}$ ⁠.

y = \frac{b}{a}\sqrt{a^2 - \frac{a^2}{2}} = \frac{b}{\sqrt2}

Why?

Put $x^2 = \frac{a^2}{2}$ in $y$ ⁠. Then $a^2 - \frac{a^2}{2} = \frac{a^2}{2}$ ⁠, whose square root is $\frac{a}{\sqrt2}$ ⁠, and $\frac{b}{a} \times \frac{a}{\sqrt2} = \frac{b}{\sqrt2}$ ⁠.

\begin{aligned} y &= \frac{b}{a}\sqrt{a^2 - \frac{a^2}{2}} \\ &= \frac{b}{a}\sqrt{\frac{2a^2 - a^2}{2}} \\ &= \frac{b}{a}\sqrt{\frac{a^2}{2}} \\ &= \frac{b}{a} \times \frac{a}{\sqrt2} = \frac{b}{\sqrt2} \end{aligned}

The length is $2x = \frac{2a}{\sqrt2} = a\sqrt2$ ⁠, since $\frac{2}{\sqrt2} = \sqrt2$ ⁠. In the same way $2y = b\sqrt2$ ⁠.

\begin{aligned} \frac{2}{\sqrt2} &= \frac{\sqrt2 \times \sqrt2}{\sqrt2} = \sqrt2 \\[4pt] 2x &= 2 \times \frac{a}{\sqrt2} = a\sqrt2 \\ 2y &= 2 \times \frac{b}{\sqrt2} = b\sqrt2 \end{aligned}

$\therefore$ For maximum area, the length of the soccer field should be $2x = a\sqrt2$ and its width should be $2y = b\sqrt2$ ⁠.

(iii) (b) Squaring both sides of $A = 2x \times 2\frac{b}{a}\sqrt{a^2 - x^2}$ ⁠, for $x \in (0, a)$ ⁠, we get

\begin{aligned} Z = A^2 &= \frac{16b^2}{a^2}x^2(a^2 - x^2) \\ &= \frac{16b^2}{a^2}(x^2a^2 - x^4) \end{aligned}

A is maximum when Z is maximum.

Why?

$A$ is positive, and for positive numbers a larger $A$ means a larger $A^2$ ⁠. So $A$ and $Z = A^2$ are maximum at the same $x$ ⁠.

Squaring removes the square root, so $Z$ is a polynomial and its second derivative is easy to find.

\begin{aligned} A^2 &= \left(\frac{4b}{a}\right)^2 x^2\left(\sqrt{a^2 - x^2}\right)^2 \\ &= \frac{16b^2}{a^2}x^2(a^2 - x^2) \\[4pt] x^2(a^2 - x^2) &= x^2a^2 - x^2 \cdot x^2 \\ &= x^2a^2 - x^4 \end{aligned}

\begin{aligned} \frac{dZ}{dx} &= \frac{16b^2}{a^2}(2xa^2 - 4x^3) \\ &= \frac{32b^2}{a^2}x\left(a + \sqrt2x\right) \\ &\qquad \times \left(a - \sqrt2x\right) \end{aligned}

\frac{dZ}{dx} = 0 \Rightarrow x = \frac{a}{\sqrt2}

Why?

Differentiate each term, with $a$ a constant:

\begin{aligned} \frac{d}{dx}(x^2a^2) &= 2x \cdot a^2 = 2xa^2 \\ \frac{d}{dx}(x^4) &= 4x^3 \end{aligned}

Take out $2x$ ⁠: $2xa^2 - 4x^3 = 2x(a^2 - 2x^2)$ ⁠, and $16 \times 2 = 32$ ⁠. Then use a difference of squares:

\begin{aligned} a^2 - 2x^2 &= a^2 - (\sqrt2x)^2 \\ &= (a + \sqrt2x)(a - \sqrt2x) \end{aligned}

Then $a - \sqrt2x = 0$ gives $\sqrt2x = a$ ⁠, so $x = \frac{a}{\sqrt2}$ ⁠.

For $0 < x < a$ only the last factor can be zero, at $x = \frac{a}{\sqrt2}$ ⁠.

\frac{d^2Z}{dx^2} = \frac{32b^2}{a^2}(a^2 - 6x^2)

Why?

Differentiate $2xa^2 - 4x^3$ again: $2a^2 - 12x^2 = 2(a^2 - 6x^2)$ ⁠. With the constant $\frac{16b^2}{a^2}$ ⁠, this gives $\frac{32b^2}{a^2}$ ⁠.

\begin{aligned} \frac{d}{dx}(2xa^2) &= 2a^2 \\ \frac{d}{dx}(4x^3) &= 4 \times 3x^2 = 12x^2 \\[4pt] \frac{16b^2}{a^2} \times 2(a^2 - 6x^2) &= \frac{32b^2}{a^2}(a^2 - 6x^2) \end{aligned}

At $x = \frac{a}{\sqrt2}$ ⁠,

\begin{aligned} \frac{d^2Z}{dx^2} &= \frac{32b^2}{a^2}(a^2 - 3a^2) \\ &= -64b^2 < 0 \end{aligned}

Why?

With $x^2 = \frac{a^2}{2}$ ⁠, $6x^2 = 6 \times \frac{a^2}{2} = 3a^2$ ⁠. Then $a^2 - 3a^2 = -2a^2$ ⁠, and

\frac{32b^2}{a^2} \times (-2a^2) = -64b^2

Second derivative test: $Z' = 0$ and $Z'' < 0$ give a local maximum.

Hence, by the second derivative test, there is a local maximum value of Z at the critical point $x = \frac{a}{\sqrt2}$ ⁠. Since there is only one critical point, Z is maximum at $x = \frac{a}{\sqrt2}$ ⁠; hence, A is maximum at $x = \frac{a}{\sqrt2}$ ⁠, where $y = \frac{b}{\sqrt2}$ ⁠.

$\therefore$ For maximum area, the length of the soccer field should be $2x = a\sqrt2$ and its width should be $2y = b\sqrt2$ ⁠.

Where this comes from

Chapter 6, Application of Derivatives: Maximum area: rectangle inscribed in an ellipse.

Question type: Word problems

How to spot it: click to see

Type 10. Word problems: the biggest box, the least cost

Spot it: a real object and the words "maximum", "minimum", "largest" or "least".

Here: With the corner $(x, y)$ on the ellipse, the area in $x$ alone is $A = \frac{4b}{a}x\sqrt{a^2 - x^2}$ for $x$ in $(0, a)$ ⁠. Its only critical point is $x = \frac{a}{\sqrt2}$ ⁠: (iii)(a) checks the sign of $A'$ around it, (iii)(b) uses $Z = A^2$ for the second derivative.

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