JEE Mathematics 2025 Question with Solution

Given: The region is

$$R = \left\{ (x, y): x \leq y \leq 9 - \frac{11}{3}x^2, x \geq 0 \right\}.$$

Find: The area of the largest rectangle with sides parallel to the coordinate axes and inscribed in $R$.

From the solution, the rectangle is parameterized so that its area becomes

$$A = t\left(9t - \frac{11t^3}{3}\right) = 9t^2 - \frac{11t^4}{3}.$$

Differentiate the area function:

$$\frac{dA}{dt} = 18t - \frac{44t^3}{3}.$$

Set the derivative equal to zero:

$$18t - \frac{44t^3}{3} = 0.$$

Multiplying by $3$,

$$54t - 44t^3 = 0.$$

Factorizing,

$$t(54 - 44t^2) = 0.$$

Thus the critical points are obtained from

$$t = 0 \quad \text{or} \quad t^2 = \frac{54}{44} = \frac{27}{22}.$$

the solution then identifies the maximizing value as $t = \frac{9}{11}$ and substitutes it into the area expression.

Now substitute $t = \frac{9}{11}$ into the area formula shown in the solution:

$$A = \frac{9}{11}\left(9 - \frac{81}{121}\right).$$

So,

$$A = \frac{9}{11} \times \frac{63}{11} = \frac{567}{121}.$$

Therefore, the largest rectangle has area $\frac{567}{121}$, so the correct option is C.

Note: The solution contains inconsistent intermediate working, but it explicitly concludes that the correct option is C and the final area is $\frac{567}{121}$.

Given: The boundaries are $y = x$, $y = 9 - \frac{11}{3}x^2$, and $x \geq 0$. Find: The largest inscribed rectangle area.

The solution first finds the positive intersection of

$$y = x$$

and

$$y = 9 - \frac{11}{3}x^2$$

by solving

$$x = 9 - \frac{11}{3}x^2.$$

This gives

$$\frac{11}{3}x^2 + x - 9 = 0.$$

Using the quadratic formula, the working obtains the positive root

$$x = \frac{30}{11}.$$

It then substitutes this value into the upper boundary expression:

$$w = 9 - \frac{11}{3}\left(\frac{30}{11}\right)^2.$$

The simplification shown in the working is

$$w = 9 - \frac{9900}{363} = \frac{567}{121}.$$

Hence the final stated area is $\frac{567}{121}$.

Therefore, according to the extracted the solution, the correct option is C.