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
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.
2 answers
where does the 6 come from