Question

A rectangle is inscribed in the parabola \[ y=9-x^{2} \] such that two of its vertices are on the X-axis and another two on the parabola. The dimensions of such rectangle lying above the X-axis and having maximum area is:

• A. \(6,\;5\sqrt3\)
• B. \(6,\;2\sqrt3\)
• C. \(9,\;6\)
• D. \(5,\;3\)

Hint:

For rectangles inscribed in symmetric curves, use symmetry to reduce the problem to a single variable optimization problem.

Answer 1

The Correct Option is B

Concept: To maximize the area of a rectangle inscribed in a parabola, we express the area as a function of a single variable and then use differentiation. Since the parabola is symmetric about the Y-axis, the rectangle with maximum area will also be symmetric about the Y-axis.

Step 1: Assume coordinates of the upper vertices.
Let the upper vertices of the rectangle be \[ (x,y)\quad\text{and}\quad(-x,y). \] Since these points lie on \[ y=9-x^2, \] we have \[ y=9-x^2. \]

Step 2: Find length and breadth of the rectangle.
Width of the rectangle \[ =2x. \] Height of the rectangle \[ =9-x^2. \] Therefore area \[ A(x)=2x(9-x^2). \] \[ A(x)=18x-2x^3. \]

Step 3: Differentiate the area function.
\[ A'(x)=18-6x^2. \] For maximum area, \[ A'(x)=0. \] \[ 18-6x^2=0. \] \[ x^2=3. \] \[ x=\sqrt3. \]

Step 4: Verify maximum area condition.
\[ A''(x)=-12x. \] At \[ x=\sqrt3, \] \[ A''(\sqrt3)<0. \] Hence area is maximum.

Step 5: Find the dimensions.
Width \[ =2\sqrt3. \] Height \[ =9-3=6. \] Therefore dimensions are \[ \boxed{6,\;2\sqrt3}. \]