# A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola #y=6-x^2#. What are the dimensions of such a rectangle with the greatest possible area? thanks for any help!?

Height:

Width

Now you differentiate to find the maximum.

Hopefully this helps!

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The dimensions of the rectangle with the greatest possible area are: Width: 2√6 units Height: 12 units

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To find the dimensions of the rectangle with the greatest possible area, we first need to find the coordinates of the two points where the rectangle touches the parabola.

- The parabola equation is y = 6 - x^2.
- Since the base of the rectangle is on the x-axis, the y-coordinate of the points where the rectangle touches the parabola will be 0.
- Set y = 0 in the parabola equation: 0 = 6 - x^2.
- Solve for x: x^2 = 6.
- Take the square root of both sides: x = ±√6.

So, the x-coordinates of the two points where the rectangle touches the parabola are ±√6.

The width of the rectangle is the difference between these x-coordinates: 2√6.

To find the height of the rectangle, substitute one of the x-values (±√6) into the parabola equation:

When x = √6: y = 6 - (√6)^2 = 6 - 6 = 0.

When x = -√6: y = 6 - (-√6)^2 = 6 - 6 = 0.

Since the y-coordinate of both points is 0, the height of the rectangle is also 0.

Therefore, the dimensions of the rectangle with the greatest possible area are: Width = 2√6 Height = 0

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