1. Set up the rectangle inside the ellipse

   Consider a rectangle centered at the origin, with one vertex in the first quadrant at $(x,y)$.
   Then the full rectangle has dimensions: $2x \times 2y$ so its area is $A=4xy.$

   Since the vertex lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ we can express $y$ in terms of $x$: $y=b\sqrt{1-\frac{x^2}{a^2}}.$

2. Write area as a function of $x$

   Substitute into $A=4xy$: $A(x)=4x\,b\sqrt{1-\frac{x^2}{a^2}}.$

3. Differentiate to maximize area

   It is easier to maximize $A^2$ or equivalently maximize $x^2y^2$. Another standard way is: $A(x)=4bx\sqrt{1-\frac{x^2}{a^2}}.$

   Let $f(x)=x\sqrt{1-\frac{x^2}{a^2}}.$

   Differentiate: $f'(x)=\sqrt{1-\frac{x^2}{a^2}}+x\left(\frac{1}{2\sqrt{1-x^2/a^2}}\right)\left(-\frac{2x}{a^2}\right).$

   So, $f'(x)=\sqrt{1-\frac{x^2}{a^2}}-\frac{x^2/a^2}{\sqrt{1-x^2/a^2}}.$

   Setting $f'(x)=0$: $\sqrt{1-\frac{x^2}{a^2}}=\frac{x^2/a^2}{\sqrt{1-x^2/a^2}}.$

   Multiply through by $\sqrt{1-x^2/a^2}$: $1-\frac{x^2}{a^2}=\frac{x^2}{a^2}.$

   Hence, $1=2\frac{x^2}{a^2}$ $x^2=\frac{a^2}{2}$ $x=\frac{a}{\sqrt2}.$

4. Find corresponding $y$

   From the ellipse equation: $\frac{x^2}{a^2}=\frac{1}{2}$ so $\frac{y^2}{b^2}=1-\frac{1}{2}=\frac{1}{2}.$

   Therefore, $y=\frac{b}{\sqrt2}.$

5. Compute maximum area

   $A_{\max}=4xy=4\left(\frac{a}{\sqrt2}\right)\left(\frac{b}{\sqrt2}\right)=4\cdot \frac{ab}{2}=2ab.$

6. Evaluate options

   • A: $2ab$ ✅
   • B: $ab$ ❌
   • C: $\sqrt{ab}$ ❌
   • D: $\dfrac{a}{b}$ ❌

Therefore, the area of the greatest rectangle inscribed in the ellipse is $\boxed{2ab}.$