CBSE Class 12 Mathematics: Application of Derivatives (Maxima & Minima)
Step 1 (Formula & Given Data): [1 Mark]
Let the height of the first pole be $h_1 = 22\text{ m}$ and the second pole be $h_2 = 16\text{ m}$. The width of the road is $20\text{ m}$. Point R is at a distance $x$ from the $16\text{ m}$ pole, so it is at a distance $(20-x)$ from the $22\text{ m}$ pole. By Pythagoras theorem, the lengths of the ladders are:
$$l_1 = \sqrt{x^2 + 16^2} = \sqrt{x^2 + 256}$$ $$l_2 = \sqrt{(20-x)^2 + 22^2} = \sqrt{(20-x)^2 + 484}$$Step 2 (Step-by-Step Mathematical Working): [2 Marks]
(i) Express $p(x) = l_1 + l_2$:
$$p(x) = \sqrt{x^2 + 256} + \sqrt{(20-x)^2 + 484}$$(ii) Find $p'(x)$:
Differentiating with respect to $x$ using the chain rule:
$$p'(x) = \frac{1}{2\sqrt{x^2 + 256}} \cdot (2x) + \frac{1}{2\sqrt{(20-x)^2 + 484}} \cdot 2(20-x)(-1)$$ $$p'(x) = \frac{x}{\sqrt{x^2 + 256}} - \frac{20-x}{\sqrt{(20-x)^2 + 484}}$$Step 3 (Final Concluded Answer): [2 Marks]
(iii) (a) Find $x$ for which $l_1^2 + l_2^2$ is minimum:
Let $f(x) = l_1^2 + l_2^2 = x^2 + 256 + (20-x)^2 + 484$
$$f(x) = x^2 + 256 + 400 - 40x + x^2 + 484 = 2x^2 - 40x + 1140$$For minima, $f'(x) = 0$:
$$f'(x) = 4x - 40 = 0 \implies x = 10\text{ m}$$Since $f''(x) = 4 > 0$, the function is minimum at $x = 10\text{ m}$.
OR (iii) (b):
If both poles are $16\text{ m}$, then $f(x) = x^2 + 16^2 + (20-x)^2 + 16^2$.
$$f'(x) = 2x - 2(20-x) = 4x - 40 = 0 \implies x = 10\text{ m}$$The ladder should be placed at $10\text{ m}$ from each pole (the midpoint of the road).
💡 EL10 Concept Insight: Imagine you are trying to connect two points with the