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?
Show 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.
• Step 1 : Understanding the Question:
We need to find the maximum possible area of a rectangle inscribed under the parabola \(y = 1 - x^2\) and above the \(X\)-axis.
The sides of the rectangle are parallel to the coordinate axes.
• Step 2 : Key Formula or Approach:
By symmetry, the vertical sides of the rectangle will be at some \(x\) and \(-x\) for \(x \in (0, 1)\).
The width of this rectangle is \(2x\).
The height of the rectangle is given by the \(y\)-coordinate of the curve at \(x\)., which is \(y = 1 - x^2\).
The area of the rectangle is \(A(x) = \text{width} \times \text{height} = 2x(1 - x^2)\).
To find the maximum area, we differentiate \(A(x)\) with respect to \(x\)., find the critical points, and apply the second derivative test.
• Step 3 : Detailed Explanation:
Let the width of the rectangle extend from \(-x\) to \(x\)., where \(0 < x < 1\).
Thus, the length of the base of the rectangle is \(2x\).
The height of the rectangle is:
\[ y = 1 - x^2 \]
The area function \(A(x)\) is given by:
\[ A(x) = 2x(1 - x^2) = 2x - 2x^3 \]
We differentiate \(A(x)\) with respect to \(x\) to find the critical points:
\[ A'(x) = \frac{d}{dx} (2x - 2x^3) = 2 - 6x^2 \]
Set the derivative to zero:
\[ 2 - 6x^2 = 0 \implies 6x^2 = 2 \implies x^2 = \frac{1}{3} \]
Since \(x\) must be positive, we take:
\[ x = \frac{1}{\sqrt{3}} \]
Let us apply the second derivative test to confirm this is a local maximum:
\[ A''(x) = -12x \]
Evaluating at \(x = \frac{1}{\sqrt{3}}\):
\[ A''\left(\frac{1}{\sqrt{3}}\right) = -12\left(\frac{1}{\sqrt{3}}\right) = -4\sqrt{3} < 0 \]
Since the second derivative is negative, the area is indeed maximized at \(x = \frac{1}{\sqrt{3}}\).
Now, let us calculate the maximum area:
\[ A_{\text{max}} = 2\left(\frac{1}{\sqrt{3}}\right) \left(1 - \left(\frac{1}{\sqrt{3}}\right)^2\right) \]
\[ A_{\text{max}} = \frac{2}{\sqrt{3}} \left(1 - \frac{1}{3}\right) = \frac{2}{\sqrt{3}} \cdot \frac{2}{3} = \frac{4}{3\sqrt{3}} \]
• Step 4 : Final Answer:
The largest area of the inscribed rectangle is \(\frac{4}{3\sqrt{3}}\).
This corresponds to option (A).
