Posted by **Mary** on Tuesday, November 4, 2008 at 5:29pm.

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=9-x^2. What are the dimensions of such a rectangle with the greatest possible area?

- Math -
**Reiny**, Tuesday, November 4, 2008 at 6:32pm
let the point of contact in the first quadrant be (x,y)

then the base of the rectangle is 2x and its height is y

Area = 2xy

= 2x(9-x^2)

= 18x - 2x^2

d(area)/dx = 18 - 4x

so 18-4x=0

x = ....

take it from here, let me know what you got

- Math -
**Damon**, Tuesday, November 4, 2008 at 6:38pm
Well, this is an upside down parabola (sheds water) with x intercepts at x = -3 and x = +3 and vertex at (0,9)

so lets do it just for positive x and double the base at the end.

Area = x y

where y = 9-x^2

A = 9x - x^3

dA/dx = 9 - 3 x^2

that is zero when x^2 = 3

or x = +/- sqrt 3 use + sqrt 3

so y = 9 - 3 = 6

Now we double the base because we only did the right half

base = 2 sqrt 3

height = 6

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