A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y = 12 - x². What are the dimensions of such a rectangle with the greatest possible area?

Solution:

Given, y = 12 - x² is an even function.

Therefore, its rectangle form also is even at the origin.

We know area of rectangle = length × width

Here, length = 2x

Width = y

Area, A = 2x (12 - x²)

A = 24x - 2x³

Take derivative of A, we get

A’ = 24 - 6x²

The area is largest when A’ = 0

⇒ 24 - 6x² = 0

⇒ x² = 4

⇒ x = 2

Put the value of x in y = 12 - x²

⇒ y = 12 - 4 = 8

Area = 2(2)(8) = 32

Therefore, the dimensions are length 2 and breath 8.

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A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y = 12 - x². What are the dimensions of such a rectangle with the greatest possible area?

Summary:

A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y = 12 - x². The dimensions of such a rectangle with the greatest possible area are length 2 and breadth 8.