CBSE 2025, Set 1

Question

**Case Study - 2**
A ladder of fixed length 'h' is to be placed along the wall such that it is free to move along the height of the wall.
![Diagram Ladder](https://www.edvaya.com/images/cbse/mathematics/2025/set1/q37-ladder.webp)

Based upon the above information, answer the following question:
(iii) (a) Show that the area ($A$) of the right triangle is maximum at the critical point.

Edvaya · Accepted Answer

The correct answer is Subjective. We have $Z = A^2$ and its first derivative $\frac{dZ}{dx} = \frac{1}{2}(h^2x - 2x^3)$. We found the critical point at $x^2 = \frac{h^2}{2}$. To show that the area is maximum, we use the second derivative test. Differentiate $\frac{dZ}{dx}$ again with respect to $x$: $\frac{d^2Z}{dx^2} = \frac{1}{2} \cdot \frac{d}{dx}(h^2x - 2x^3)$ $\frac{d^2Z}{dx^2} = \frac{1}{2}(h^2 - 6x^2)$ Now, evaluate the second derivative at the critical point $x^2 = \frac{h^2}{2}$: $\frac{d^2Z}{dx^2} = \frac{1}{2} [h^2 - 6(\frac{h^2}{2})]$ $\frac{d^2Z}{dx^2} = \frac{1}{2} [h^2 - 3h^2] = \frac{1}{2} [-2h^2] = -h^2$ Since $h$ is a positive length, $-h^2 < 0$. Because the second derivative is strictly negative at the critical point, $Z$ (and therefore the Area $A$) achieves a **local maximum** at $x = \frac{h}{\sqrt{2}}$.

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Mathematics > Calculus > Application Of Derivatives

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