1. Understand the region

The region is

So the region is bounded by:

• the line $y=x$,
• the parabola $y=9-\frac{11}{3}x^2$,
• and $x\ge 0$.

We need the largest rectangle with sides parallel to the coordinate axes inscribed in this region.

2. Geometry of the rectangle

Let the rectangle have vertical sides at $x=0$ and $x=a$.

For the rectangle to lie entirely inside the region:

• its bottom side must be at or above $y=x$ for all $x\in[0,a]$,
• its top side must be at or below $y=9-\frac{11}{3}x^2$ for all $x\in[0,a]$.

Since $y=x$ is increasing, the largest possible constant bottom level is determined by the worst point $x=a$, hence

Since $9-\frac{11}{3}x^2$ is decreasing for $x\ge 0$, the largest possible constant top level is also determined by $x=a$, hence

Thus the rectangle has:

• width $=a$,
• height $=\left(9-\frac{11}{3}a^2\right)-a$.

So its area is

3. Find the critical point

Differentiate:

Set $A'(a)=0$:

Solve:

So,

or $a=-1$ (reject since $a\ge 0$).

Hence the only relevant critical point is

4. Check that it gives maximum area

Second derivative:

At $a=\frac{9}{11}$,

So this gives a maximum.

5. Compute the maximum area

Substitute $a=\frac{9}{11}$ into $A(a)$:

Now simplify term by term:

Therefore,

Convert $\frac{81}{11}$ to denominator $121$:

So,

6. Match with the options

is Option B.

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.