Question

A rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, then its dimensions are

• A. $4\sqrt{2}$, $6\sqrt{2}$
• B. $2$, $5\sqrt{2}$
• C. $4\sqrt{2}$, $5\sqrt{2}$
• D. $4\sqrt{2}$, $2$

Hint:

For any standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the sides of the maximum inscribed rectangle are always fixed structurally as $\sqrt{2}a$ and $\sqrt{2}b$. Knowing this shortcut allows you to calculate the answers directly: $\sqrt{2}(5) = 5\sqrt{2}$ and $\sqrt{2}(4) = 4\sqrt{2}$ within seconds!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the geometric dimensions (length and width) of the specific rectangle with the maximum possible area that can be perfectly inscribed inside the standard ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Step 2: Key Formula or Approach:
For a standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, any point on the ellipse can be parametrically represented as $(a\cos\theta, b\sin\theta)$. An inscribed rectangle's vertices will lie in the four quadrants at $(\pm a\cos\theta, \pm b\sin\theta)$. The total length of the rectangle is $2a\cos\theta$. The total breadth of the rectangle is $2b\sin\theta$. The area $A = (2a\cos\theta)(2b\sin\theta) = 2ab\sin(2\theta)$. The area is maximized when $\sin(2\theta) = 1$, which occurs at $2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}$.

Step 3: Detailed Explanation:
From the given ellipse equation: $$a^2 = 25 \implies a = 5$$ $$b^2 = 16 \implies b = 4$$ The parametric dimensions for the inscribed rectangle are: $$\text{Dimension 1} = 2a\cos\theta$$ $$\text{Dimension 2} = 2b\sin\theta$$ For maximum area, substitute $\theta = \frac{\pi}{4}$: $$\text{Dimension 1} = 2(5)\cos\left(\frac{\pi}{4}\right) = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}$$ $$\text{Dimension 2} = 2(4)\sin\left(\frac{\pi}{4}\right) = 8 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}$$ Thus, the maximum dimensions are $5\sqrt{2}$ and $4\sqrt{2}$.

Step 4: Final Answer:
The dimensions of the maximum area rectangle are $4\sqrt{2}$ and $5\sqrt{2}$, which matches option (C).