1. Understand the region

The parabola is $y^2=2x \implies x=\frac{y^2}{2}.$

The line is $x=24.$

So the region enclosed by these curves lies between the parabola and the vertical line $x=24$.

Since the parabola is symmetric about the $x$-axis, the rectangle of maximum area will also be symmetric about the $x$-axis.

2. Take a general inscribed rectangle

Let the upper and lower vertices on the line $x=24$ be at heights $y$ and $-y$.

Then the corresponding points on the parabola are at $x=\frac{y^2}{2}.$

So for the rectangle:

• height $=2y$
• width $=24-\frac{y^2}{2}$

Hence area is $A(y)=2y\left(24-\frac{y^2}{2}\right)=48y-y^3.$

3. Maximize the area

Differentiate: $A'(y)=48-3y^2.$

Set equal to zero: $48-3y^2=0$ $3y^2=48$ $y^2=16$ $y=4$ (We take $y>0$.)

Now check second derivative: $A''(y)=-6y.$ At $y=4$, $A''(4)=-24<0,$ so the area is maximum.

4. Compute the maximum area

Substitute $y=4$ into $A(y)$: $A_{\max}=48(4)-4^3=192-64=128.$

5. Compare with stored answer

Derived answer: $128$

Stored correct answer: $128$

They match.