What are the dimensions of the rectangle with the greatest area that can be inscribed beneath the parabola y = 12 – x^2, with its base on the x-axis?

To find the dimensions of the rectangle with the greatest possible area that can be inscribed under the parabola defined by the equation y = 12 – x², with its base on the x-axis, we can follow these steps:

1. Understand the Geometry:

We need to inscribe a rectangle in such a way that the upper corners touch the parabola while the base rests on the x-axis. If we let ‘x’ be half the width of the rectangle, the coordinates of the upper corners of the rectangle will be (-x, y) and (x, y), where ‘y’ is the height of the rectangle. According to the parabola’s equation, the height at any x is given by:

y = 12 – x²

2. Define the Area of the Rectangle:

The area ‘A’ of the rectangle can be expressed as:

A = width × height = (2x) × y = 2x(12 – x²)

So the area function becomes:

A(x) = 2x(12 – x²) = 24x – 2x³

3. Optimize the Area:

To find the maximum area, we need to take the derivative of the area function and set it to zero:

A'(x) = 24 – 6x²

To find the critical points, set the derivative equal to zero:

24 – 6x² = 0

6x² = 24

x² = 4

x = 2 or x = -2 (we only consider x = 2 because we are dealing with the positive width).

4. Calculate Maximum Area:

Next, we calculate the height of the rectangle when ‘x’ is 2:

y = 12 – x² = 12 – 2² = 12 – 4 = 8

5. Determine the Dimensions:

The rectangle dimensions can now be defined:

  • Width: 2 * x = 2 * 2 = 4
  • Height: y = 8

Final Result:

The dimensions of the rectangle with the greatest possible area that can be inscribed under the parabola are:

  • Width: 4 units
  • Height: 8 units

This gives us a maximum area of:

Area = Width × Height = 4 × 8 = 32 square units.

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