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?
Width =

Height =

Height: 4 units

Width 2√/2

Explanation:

Start by sketching y = 6 - ². Then draw a rectangle beneath it. You will notice that the width is 2x and the height is 6 ². Area is given by length - times width, so the area function will be A=2x (6 - x²) = 12x - 2x³.

Now you differentiate to find the maximum.

A' = 12 - 6x²

Find critical numbers by setting A' to 0.

x = ± √/2

The derivative is negative at x = 2 and positive at x = 1, which justifies that the rectangle with width of √2 has maximal area.

The height will be y (√2) = 6 – (√2)²) = 4

4 and 5/12ft.

Step-by-step explanation:

you need to convert 4 and 3/4 into 19/4. now you need the 2 numbers to have a common denominator.

multiply the top and bottom number of 19/4 by 3 and get 57/12

multiply the top and bottom number of 1/3 by 4 and get 4/12

subtract 4/12 from 57/12 and get 53/12

12 can go into 53 4 times. so you get 4and 5/12 ft.

hop this helps

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