Question

What is the largest area of a rectangle, whose sides are parallel to the coordinate axes, that can be inscribed under the graph of the curve
$y = 1 - x^2$
and above the $X$-axis?

• A. $\frac{4}{3\sqrt{3}}$
• B. $\frac{2}{3\sqrt{3}}$
• C. $\frac{4}{3}$
• D. $\frac{1}{3}$

Hint:

For any rectangle inscribed under \(y = a - bx^2\)., the maximum area is achieved when the width is \(2x\) where \(x = \sqrt{\frac{a}{3b}}\).
Applying this shortcut directly to \(y = 1 - x^2\) gives \(x = \frac{1}{\sqrt{3}}\) instantly.

Answer 1

The Correct Option is A

To find the largest area of a rectangle, whose sides are parallel to the coordinate axes, that can be inscribed under the graph of the curve \(y = 1 - x^2\) and above the X-axis, follow these steps:

1. Understand the problem: The rectangle's top sides must lie under the curve, and it must be above the X-axis. This implies the rectangle's vertices on the curve will be \((x, 1-x^2)\) and \((-x, 1-x^2)\).
2. The width of the rectangle is \(2x\) (as it spans symmetrically from \(-x\) to \(x\)) and the height is \(1-x^2\). Therefore, the area of the rectangle \((A)\) is given by:
3. \(A = 2x(1-x^2) = 2x - 2x^3\).
4. To maximize the area, take the derivative of \(A\) with respect to \(x\) and set it to zero:
5. \(\frac{dA}{dx} = 2 - 6x^2\).
6. Set the derivative to zero for the critical points:
7. \(2 - 6x^2 = 0\)
   \(6x^2 = 2\)
   \(x^2 = \frac{1}{3}\)
   \(x = \pm \frac{1}{\sqrt{3}}\).
8. Since the rectangle spans from \(-x\) to \(x\), we use the positive value \(x = \frac{1}{\sqrt{3}}\).
9. Substitute back to find the maximum area:
10. \(A = 2\left(\frac{1}{\sqrt{3}}\right)\left(1-\left(\frac{1}{\sqrt{3}}\right)^2\right)\)
    \(= 2 \cdot \frac{1}{\sqrt{3}} \cdot \left(1-\frac{1}{3}\right)\)
    \(= 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{3}\)
    \(= \frac{4}{3\sqrt{3}}\).

Therefore, the largest area of the rectangle that can be inscribed under the graph and above the X-axis is \(\frac{4}{3\sqrt{3}}\).