Case Study Questions (AOD)
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2026 Board (Sample)
00:00
A company produces bulbs and the profit function is given by $P(x) = -x^2 + 28x - 120$ (where $x$ is the number of bulbs in thousands). (i) Find the number of bulbs for which profit is maximum. (ii) Find the maximum profit. (iii) Find the intervals where profit is increasing. 4 Marks
i) $P'(x) = -2x + 28$. Set to zero: $x = 14$ (14,000 bulbs).
ii) $P(14) = -(196) + 28(14) - 120 = -196 + 392 - 120 = 76$. Max Profit = 76 units.
iii) $P'(x) > 0 \Rightarrow -2x + 28 > 0 \Rightarrow x < 14$. Increasing in $(0, 14)$.
x=14, Max Profit=76
Q2
2023
00:00
A ball is thrown upward and its height $h$ in meters at time $t$ seconds is $h(t) = 20t - 5t^2$. (i) Find the maximum height reached by the ball. (ii) Find the velocity of the ball at $t = 1$ second. 4 Marks
i) $h'(t) = 20 - 10t$. Set to zero: $t = 2$ s.
$h(2) = 20(2) - 5(4) = 40 - 20 = 20$ m.
ii) $v(t) = h'(t) = 20 - 10t$. At $t = 1$, $v(1) = 10$ m/s.
Max Height: 20m, Velocity at 1s: 10m/s
Q3
2025 Board
00:00
A manufacturer can sell $x$ items at a price of $p = 5 - x/100$ each. The cost price of $x$ items is $C(x) = x/5 + 500$. (i) Find the revenue function $R(x)$. (ii) Find the number of items for maximum profit. 4 Marks
i) Revenue $R(x) = p \cdot x = x(5 - x/100) = 5x - x^2/100$.
ii) Profit $P(x) = R(x) - C(x) = (5x - x^2/100) - (x/5 + 500)$.
$P'(x) = 5 - x/50 - 1/5 = 4.8 - x/50$.
Set $P'(x) = 0 \Rightarrow x/50 = 4.8 \Rightarrow x = 240$.
R(x) = 5x - x²/100, x = 240
Q4
2024 Board
00:00
A farmer wants to fence a rectangular field with 400m of fencing. One side is a river. (i) Write area $A$ as a function of $x$. (ii) Find dimensions for maximum area. 4 Marks
i) $2x + y = 400 \Rightarrow y = 400 - 2x$. Area $A(x) = x(400-2x) = 400x - 2x^2$.
ii) $A'(x) = 400 - 4x = 0 \Rightarrow x = 100$. $y = 200$.
Dimensions: 100m × 200m.
A(x) = 400x - 2x², 100m × 200m
Q5
2022 Board
00:00
A cuboidal box with a square base and open top is to be made of 48 sq. units of cardboard. (i) Express height $h$ in terms of base side $x$. (ii) Find dimensions for maximum volume. 4 Marks
i) Surface Area $S = x^2 + 4xh = 48 \Rightarrow 4xh = 48 - x^2 \Rightarrow h = \frac{48-x^2}{4x}$.
ii) Volume $V = x^2h = x^2 \cdot \frac{48-x^2}{4x} = \frac{1}{4}(48x - x^3)$.
$V'(x) = \frac{1}{4}(48 - 3x^2) = 0 \Rightarrow 3x^2 = 48 \Rightarrow x^2 = 16 \Rightarrow x = 4$.
$h = (48-16)/16 = 2$. Dimensions: 4 × 4 × 2.
h = (48-x²)/4x, 4 × 4 × 2