What is the largest area of a rectangle that can be inscribed under the parabola y = 12 – x², with its base on the x-axis?

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To find the largest area of a rectangle inscribed under the parabola given by the equation y = 12 – x² and with its base on the x-axis, we can follow these steps:

  1. First, we recognize that the rectangle has its base on the x-axis, and its upper corners will touch the parabola. Since the parabola is symmetric, we can denote one corner of the rectangle as
    (x, y) while the other will be at (-x, y), making the width of the rectangle 2x.

  2. Next, we establish the height of the rectangle, which is determined by the y-coordinate of the points on the parabola. Thus, we can express the height as:

    height = y = 12 - x²

  3. The area A of the rectangle can therefore be expressed as a function of x:

    A = width × height = 2x(12 - x²)

    This simplifies to:

    A = 24x - 2x³

  4. To find the maximum area, we need to differentiate this area function with respect to x and then set the first derivative A' to zero:

    A' = 24 - 6x² = 0

    This gives us:

    6x² = 24

    which simplifies to:

    x² = 4

    So, taking the square root:

    x = 2 (since negative values don’t apply to this context).

  5. Now we can find the corresponding height:

    y = 12 - (2)² = 12 - 4 = 8

  6. Hence, the dimensions of the rectangle when x = 2 will be a width of 2x = 4 and a height of 8.

  7. The maximum area can be calculated now:

    Area = width × height = 4 × 8 = 32

Therefore, the largest area of the rectangle that can be inscribed under the parabola y = 12 – x² with its base on the x-axis is 32 square units.

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