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:
(ii) Find the derivative of the area ($A$) with respect to the height on the wall ($x$), and find its critical point.

Edvaya · Accepted Answer

The correct answer is Subjective. The area is $A = \frac{1}{2}x\sqrt{h^2 - x^2}$. To make differentiation easier, let us define a new function $Z = A^2$. The critical points will be the same. $Z = (\frac{1}{2}x\sqrt{h^2 - x^2})^2$ $Z = \frac{1}{4} x^2(h^2 - x^2) = \frac{1}{4}(h^2 x^2 - x^4)$ Differentiate $Z$ with respect to $x$: $\frac{dZ}{dx} = \frac{1}{4}(2h^2x - 4x^3) = \frac{1}{2}(h^2x - 2x^3)$ Alternatively, differentiating $A$ directly using the product rule: $\frac{dA}{dx} = \frac{1}{2} [1 \cdot \sqrt{h^2-x^2} + x \cdot \frac{-2x}{2\sqrt{h^2-x^2}}]$ $\frac{dA}{dx} = \frac{1}{2} [\sqrt{h^2-x^2} - \frac{x^2}{\sqrt{h^2-x^2}}]$ $\frac{dA}{dx} = \frac{1}{2} [\frac{h^2-x^2-x^2}{\sqrt{h^2-x^2}}] = \frac{h^2 - 2x^2}{2\sqrt{h^2-x^2}}$ To find the critical point, set the derivative to zero ($\frac{dA}{dx} = 0$): $h^2 - 2x^2 = 0$ $2x^2 = h^2 \implies x^2 = \frac{h^2}{2}$ Since $x$ represents height, $x > 0$: $x = \frac{h}{\sqrt{2}}$

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Mathematics > Calculus > Application Of Derivatives

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy