Posted by **Isaac** on Monday, November 28, 2011 at 12:22pm.

Hello, could someone please help me with this problem? I'm a little stuck with it. Thanks, Isaac

A rectangle is to be inscribed with its base on the x-axis and its other two vertices above the x-axis on the parabola y=9-x^2

(a) find the dimensions of the rectangle of largest area.

(b) Find the area of the largest rectangle.

- Calculus -
**Steve**, Monday, November 28, 2011 at 1:18pm
Since the parabola is symmetric about the line x=0, let the rectangle have corners

(-x,0) (x,0) (x,y) (-x,y)

Since y is 9-x^2, the rectangle is

2x by (9-x^2) in width and height. The area is thus

A = 2x(9-x^2) = 18x - 2x^3

To find the value of x which gives maximum area, we want

A' = 0

A' = 18 - 6x^2

A' = 0 when x = √3

So, the rectangle is

2√3 by 6 with area 12√3

- Calculus -
**Anonymous**, Sunday, October 5, 2014 at 6:59am
where does the 6 come from

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