CBSE Sample Paper 2020-21 · Part A: Section II · 4 marks

Q17: Case study: a floor with semicircular ends and perimeter 200 m

Sample 2020-21Ch 6 · Application of Derivatives

The question

An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200m as shown below:

Design of Floor: the rectangular region A, of length

x

and breadth

y

⁠, with a semicircle of radius

\frac{y}{2}

at each end.

Based on the above information answer the following:

(i) If x and y represents the length and breadth of the rectangular region, then the relation between the variables is (1 mark)

a) $x + \pi y = 100$

b) $2x + \pi y = 200$

c) $\pi x + y = 50$

d) $x + y = 100$

(ii) The area of the rectangular region A expressed as a function of x is (1 mark)

a) $\frac{2}{\pi}(100x - x^2)$

b) $\frac{1}{\pi}(100x - x^2)$

c) $\frac{x}{\pi}(100 - x)$

d) $\pi y^2 + \frac{2}{\pi}(100x - x^2)$

(iii) The maximum value of area A is (1 mark)

a) $\frac{\pi}{3200}\ m^2$

b) $\frac{3200}{\pi}\ m^2$

c) $\frac{5000}{\pi}\ m^2$

d) $\frac{1000}{\pi}\ m^2$

(iv) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the valve of x should be (1 mark)

a) 0 m

b) 30 m

c) 50 m

d) 80 m

(v) The extra area generated if the area of the whole floor is maximized is : (1 mark)

a) $\frac{3000}{\pi}\ m^2$

b) $\frac{5000}{\pi}\ m^2$

c) $\frac{7000}{\pi}\ m^2$

d) No change Both areas are equal

Exam answer

Any 4 of the 5 sub parts are to be attempted; all 5 are answered here.

(i) The two semicircular ends, each of diameter y, make one circle of perimeter $\pi y$ ⁠.

Why?

The perimeter is the length of the outer boundary of the floor: the two straight edges of length $x$ and the two curved ends. The dashed sides of length $y$ lie inside the floor, so they are not part of it.

Each end is a semicircle of radius $\frac{y}{2}$ ⁠, so its arc has length $\pi \times \frac{y}{2}$ ⁠. Two of them give $\pi y$ ⁠.

\begin{aligned} \text{one arc} &= \tfrac12(2\pi r) = \pi r = \pi \cdot \frac{y}{2} \\ \text{two arcs} &= \frac{\pi y}{2} + \frac{\pi y}{2} = \pi y \\ \text{Perimeter} &= x + x + \pi y \\ &= 2x + \pi y \end{aligned}

\text{Perimeter} = 2x + \pi y = 200

Why?

Watch out: (d) comes from the rectangle's own perimeter $2x + 2y = 200$ ⁠, which wrongly counts the two dashed sides.

Answer: (b) $2x + \pi y = 200$

(ii) From (i), $y = \frac{200 - 2x}{\pi}$ ⁠.

Why?

A must be a function of $x$ alone, so we make $y$ the subject of the relation in (i): $\pi y = 200 - 2x$ ⁠, then divide by $\pi$ ⁠.

\begin{aligned} 2x + \pi y &= 200 \\ \pi y &= 200 - 2x \\ y &= \frac{200 - 2x}{\pi} \end{aligned}

\begin{aligned} A &= xy = \frac{x(200 - 2x)}{\pi} \\ &= \frac{2}{\pi}(100x - x^2) \end{aligned}

Why?

A is the rectangle only, so its area is length times breadth. Then take out the common factor 2:

\begin{aligned} A &= x \cdot \frac{200 - 2x}{\pi} \\ x(200 - 2x) &= 200x - 2x^2 \\ &= 2(100x - x^2) \end{aligned}

Option (d) adds an area of the ends, so it is not the rectangular region A.

Answer: (a) $\frac{2}{\pi}(100x - x^2)$

(iii)

\begin{aligned} \frac{dA}{dx} &= \frac{2}{\pi}(100 - 2x) = 0 \\ \Rightarrow x &= 50 \\ \frac{d^2A}{dx^2} &= -\frac{4}{\pi} < 0 \end{aligned}

Why?

To find a maximum, set the first derivative to 0, then check the sign of the second derivative.

The constant $\frac{2}{\pi}$ stays in front. $\frac{d}{dx}(100x - x^2) = 100 - 2x$ ⁠, which is 0 when $x = 50$ ⁠. Differentiating again gives $\frac{2}{\pi}(-2) = -\frac{4}{\pi}$ ⁠.

\begin{aligned} \frac{d}{dx}(100x) &= 100 \\ \frac{d}{dx}(x^2) &= 2x \\ \frac{dA}{dx} &= \frac{2}{\pi}(100 - 2x) \end{aligned}

Set it to 0. Since $\frac{2}{\pi} \ne 0$ ⁠, the bracket is 0:

\begin{aligned} 100 - 2x &= 0 \\ 2x &= 100 \\ x &= \frac{100}{2} = 50 \end{aligned}

Second derivative:

\begin{aligned} \frac{d}{dx}(100 - 2x) &= 0 - 2 = -2 \\ \frac{d^2A}{dx^2} &= \frac{2}{\pi}(-2) = -\frac{4}{\pi} \end{aligned}

A negative second derivative means the graph bends down, so $x = 50$ gives a maximum.

So A is maximum at $x = 50$ ⁠.

A = \frac{2}{\pi}(5000 - 2500) = \frac{5000}{\pi}

Why?

At $x = 50$ ⁠: $100x = 5000$ and $x^2 = 2500$ ⁠. Then $2 \times 2500 = 5000$ ⁠.

\begin{aligned} A &= \frac{2}{\pi}\big(100(50) - (50)^2\big) \\ &= \frac{2}{\pi}(5000 - 2500) \\ &= \frac{2}{\pi}(2500) \\ &= \frac{2 \times 2500}{\pi} = \frac{5000}{\pi} \end{aligned}

Check: at $x = 50$ ⁠, $y = \frac{200 - 100}{\pi} = \frac{100}{\pi}$ ⁠, and $xy = \frac{5000}{\pi}$ ⁠.

\begin{aligned} y &= \frac{200 - 2(50)}{\pi} = \frac{200 - 100}{\pi} \\ &= \frac{100}{\pi} \\ xy &= 50 \times \frac{100}{\pi} = \frac{5000}{\pi} \end{aligned}

Answer: (c) $\frac{5000}{\pi}\ m^2$

(iv) Area of the whole floor, with $\frac{y}{2} = \frac{100 - x}{\pi}$ ⁠:

\begin{aligned} S &= xy + \pi\left(\frac{y}{2}\right)^2 \\ &= \frac{2(100x - x^2)}{\pi} + \frac{(100 - x)^2}{\pi} \\ &= \frac{10000 - x^2}{\pi} \end{aligned}

Why?

The two semicircular ends of radius $\frac{y}{2}$ together make one full circle, of area $\pi\left(\frac{y}{2}\right)^2$ ⁠. Halving the $y$ of (ii) gives $\frac{y}{2} = \frac{100 - x}{\pi}$ ⁠, so the circle has area $\pi \cdot \frac{(100 - x)^2}{\pi^2} = \frac{(100 - x)^2}{\pi}$ ⁠.

\begin{aligned} \frac{y}{2} &= \frac{1}{2} \cdot \frac{200 - 2x}{\pi} \\ &= \frac{200 - 2x}{2\pi} = \frac{2(100 - x)}{2\pi} \\ &= \frac{100 - x}{\pi} \\[4pt] \pi\left(\frac{y}{2}\right)^2 &= \pi \cdot \frac{(100 - x)^2}{\pi^2} \\ &= \frac{(100 - x)^2}{\pi} \end{aligned}

Expand the square with $(a - b)^2 = a^2 - 2ab + b^2$ ⁠:

\begin{aligned} (100 - x)^2 &= 100^2 - 2(100)(x) + x^2 \\ &= 10000 - 200x + x^2 \end{aligned}

Over the common denominator $\pi$ ⁠, the top is

\begin{aligned} &200x - 2x^2 \\ &\quad + 10000 - 200x + x^2 \\ &= 10000 - x^2 \end{aligned}

Collect like terms: $200x - 200x = 0$ and $-2x^2 + x^2 = -x^2$ ⁠.

\frac{dS}{dx} = -\frac{2x}{\pi} < 0 \text{ for } x > 0

So S decreases as x increases, and S is maximum at $x = 0$ ⁠.

Why?

Differentiate $S = \frac{1}{\pi}(10000 - x^2)$ ⁠. The constant 10000 has derivative 0:

\begin{aligned} \frac{dS}{dx} &= \frac{1}{\pi}(0 - 2x) \\ &= -\frac{2x}{\pi} \end{aligned}

$x$ is a length and $y \ge 0$ needs $200 - 2x \ge 0$ ⁠, so $0 \le x \le 100$ ⁠. On this interval S only decreases, so its largest value is at the left end, $x = 0$ ⁠.

At $x = 0$ the straight part disappears and the floor is a full circle. For a fixed perimeter, a circle encloses the most area.

Answer: (a) 0 m

(v) Maximum area of the whole floor, from (iv), at $x = 0$ ⁠:

S = \frac{10000 - 0^2}{\pi} = \frac{10000}{\pi}

Why?

Put $x = 0$ in $S = \frac{10000 - x^2}{\pi}$ ⁠.

Check: at $x = 0$ ⁠, $y = \frac{200}{\pi}$ ⁠, so the floor is a circle of radius $\frac{100}{\pi}$ ⁠, with area $\pi\left(\frac{100}{\pi}\right)^2 = \frac{10000}{\pi}$ ⁠.

\begin{aligned} r &= \frac{y}{2} = \frac{200}{2\pi} = \frac{100}{\pi} \\ \pi r^2 &= \pi \cdot \frac{100^2}{\pi^2} \\ &= \pi \cdot \frac{10000}{\pi^2} = \frac{10000}{\pi} \end{aligned}

Maximum area of the rectangular region A, from (iii), $= \frac{5000}{\pi}$ (at $x = 50$ ⁠).

Why?

The case study compares two plans. The design of (iii) makes the rectangular region A as large as possible. The CEO's plan of (iv) makes the whole floor as large as possible. The extra area is the gain from the second plan over the first.

\begin{aligned} \text{Extra area} &= \frac{10000}{\pi} - \frac{5000}{\pi} \\ &= \frac{5000}{\pi} \end{aligned}

Why?

Both areas have the denominator $\pi$ ⁠, so subtract the numerators: $10000 - 5000 = 5000$ ⁠.

Comparing whole floors at $x = 0$ and at $x = 50$ instead: at $x = 50$ ⁠, $S = \frac{10000 - 2500}{\pi} = \frac{7500}{\pi}$ ⁠, and

\frac{10000}{\pi} - \frac{7500}{\pi} = \frac{2500}{\pi}

That is not an option, so the intended comparison is with A, as above.

Watch out: the official scheme prints (d), "No change". That cannot be right, because $\frac{10000}{\pi}$ and $\frac{5000}{\pi}$ are not equal. The correct option is (b).

Answer: (b) $\dfrac{5000}{\pi}\ m^2$

The official marking scheme gives (d) for part (v). The maximum whole-floor area $\frac{10000}{\pi}\ m^2$ exceeds the maximum rectangular area $\frac{5000}{\pi}\ m^2$ by $\frac{5000}{\pi}\ m^2$ ⁠, which is option (b).

Where this comes from

Chapter 6, Application of Derivatives: Maxima: floor with semicircular ends.

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: The perimeter $2x + \pi y = 200$ turns the area of A into $\frac{2}{\pi}(100x - x^2)$ ⁠, largest at $x = 50$ in (iii). In (iv) the whole floor $\frac{10000 - x^2}{\pi}$ has a negative derivative for every $x$ above 0, so its largest value is at the end $x = 0$ ⁠.

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