Sample 2026-27 Class 12 Maths Q37 Solution: Case study: Yamuna water quality index and investment

The question

Water is essential for life, and rivers are vital water sources across the world. The Yamuna Action Plan (YAP), one of India's largest river restoration projects, was launched to clean the Yamuna, but its effectiveness (WQI) the Water Quality Index must be measured through cost-benefit, economic and environmental analysis. Therefore, comprehensive pollutant treatment, prevention of further contamination, and community awareness are crucial for sustainable river management.

Implementing the same concept to the river Yamuna, if an initial base line WQI is 35, a quadratic expression is formed to measure the impact of investment of ₹ $x$ crore on the WQI i.e., $W(x)$ as

$W(x) = 35 + 1.6x - 0.02x^2$ ⁠, where $x$ is in Rupees in crores

Use the given $W(x)$ to answer the following questions:

(i) Find the value of $x$ at the stationary point for the function $W(x)$ ⁠. (1 mark)

(ii) Find the maximum value of the WQI using the second derivative test? (1 mark)

(iii) (a) If the investment is ₹ 35 crores, find by how much the WQI is short of its maximum achievable value. (2 marks)

OR

(iii) (b) Evaluate the WQI when the investment is ₹60 crore. Determine whether this indicates an enhancement or a diminishing return in our objective when compared with the optimum WQI. (2 marks)

Exam answer

(i)

W(x) = 35 + 1.6x - 0.02x^2

\begin{aligned} W'(x) &= 1.6 - 0.04x = 0 \\ \Rightarrow x &= 40 \end{aligned}

Why?

A stationary point is where the derivative is zero. Differentiate term by term: the constant 35 gives 0, $1.6x$ gives 1.6, and $0.02x^2$ gives $0.02 \times 2x = 0.04x$ ⁠.

Then $0.04x = 1.6$ ⁠, so $x = \frac{1.6}{0.04} = \frac{160}{4} = 40$ ⁠.

\begin{aligned} 1.6 - 0.04x &= 0 \\ 1.6 &= 0.04x \\ x &= \frac{1.6}{0.04} = \frac{1.6 \times 100}{0.04 \times 100} \\ &= \frac{160}{4} = 40 \end{aligned}

$\therefore$ The value of $x$ at the stationary point of $W(x)$ is 40 (crores).

(ii)

W''(x) = -0.04 < 0

So, by the second derivative test, $W(x)$ is maximum at $x = 40$ ⁠.

Why?

Second derivative test: at a stationary point, $W'' < 0$ means a maximum and $W'' > 0$ means a minimum.

Differentiate $W'(x) = 1.6 - 0.04x$ once more: the constant 1.6 gives 0, and $-0.04x$ gives $-0.04$ ⁠.

Here $W''(x) = -0.04$ for every $x$ ⁠, so it is negative at $x = 40$ ⁠.

\begin{aligned} W(40) &= 35 + (1.6)(40) \\ &\quad - (0.02)(40)^2 \end{aligned}

W(40) = 35 + 64 - 32 = 67

Why?

$1.6 \times 40 = 64$ ⁠, and $40^2 = 1600$ ⁠, so $0.02 \times 1600 = 32$ ⁠.

\begin{aligned} 35 + 64 &= 99 \\ 99 - 32 &= 67 \end{aligned}

Watch out: the maximum value of the WQI is $W(40)$ ⁠, not 40. The question asks for the value of W.

$\therefore$ Maximum WQI $= 67$ ⁠, when ₹ 40 crores are spent.

(iii) (a)

\begin{aligned} W(35) &= 35 + (1.6)(35) \\ &\quad - (0.02)(35)^2 \\ &= 35 + 56 - 24.5 = 66.5 \end{aligned}

Why?

$1.6 \times 35 = 56$ ⁠. Also $35^2 = 1225$ ⁠, and $0.02 \times 1225 = 24.5$ ⁠.

\begin{aligned} 35 + 56 &= 91 \\ 91 - 24.5 &= 66.5 \end{aligned}

67 - 66.5 = 0.5

Why?

"Short of its maximum" is the gap between the maximum value from (ii) and the value at ₹ 35 crores.

$\therefore$ The WQI is short of its maximum value by 0.5 units.

(iii) (b)

\begin{aligned} W(60) &= 35 + (1.6)(60) \\ &\quad - (0.02)(60)^2 \end{aligned}

W(60) = 35 + 96 - 72 = 59

Why?

$1.6 \times 60 = 96$ ⁠. Also $60^2 = 3600$ ⁠, and $0.02 \times 3600 = 72$ ⁠.

\begin{aligned} 35 + 96 &= 131 \\ 131 - 72 &= 59 \end{aligned}

59 is less than the maximum value 67 of the WQI, even though the investment has been increased from ₹ 40 crores to ₹ 60 crores.

Why?

For $x > 40$ ⁠, $W'(x) = 1.6 - 0.04x < 0$ ⁠, so W decreases. Each extra crore beyond 40 lowers the WQI. More money giving a worse result is a diminishing return.

$\therefore$ This shows a diminishing return.

Where this comes from

Chapter 6, Application of Derivatives: Maximum value: second derivative test.

Question type: Local maxima and minima

How to spot it: click to see

Type 7. Local maxima and minima: the second derivative test

Spot it: "Find the local maxima and local minima" or "the local maximum value".

Here: $W'(x) = 1.6 - 0.04x = 0$ gives the stationary point $x = 40$ ⁠, and $W''(x) = -0.04$ is negative, so the maximum WQI is $W(40) = 67$ ⁠. Part (iii) compares other investments with it.

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Q37: Case study: Yamuna water quality index and investment