M mr.burger New member Joined Jun 12, 2005 Messages 21 Jun 17, 2005 #1 What is the maximum area of a rectangle whose base lies on the x-axis and whose other two vertices lie on the graph of y= -x^(2) +4?

What is the maximum area of a rectangle whose base lies on the x-axis and whose other two vertices lie on the graph of y= -x^(2) +4?

G Gene Senior Member Joined Oct 8, 2003 Messages 1,904 Jun 18, 2005 #2 the graph is symetrical about the y axis and is a downward opening parabola. If one point is (-a,y) the other top point is (a,y) and the area is A = 2a*y = 2a*(-a^2+4) = -2a^3+8a Find dA/da = 0

the graph is symetrical about the y axis and is a downward opening parabola. If one point is (-a,y) the other top point is (a,y) and the area is A = 2a*y = 2a*(-a^2+4) = -2a^3+8a Find dA/da = 0