Question

Applications of derivatives

A rectangle has its base on the x axis and its upper two vertices on the parabola y=12 - x^2.
What is the largest area the rectangle can have, and what are its dimensions.
Support your answer graphically.
Thanks.

Well, the parabola is symettric about x=0, so just solve the half of rectangle, with x=0 being the one side, and x the other.

Area= xy= x(12-x^2)= 12x-x^3

dA/dx= 12 -3x^2 or setting to zero, x=2

Now applying symettry, the rectangle is from x=-2 to x=2

Area= INt ydx from x=-2 to 2
INT (12-x^2) dx= 12x-1/3 x^3 limits..
Area= 12*2 - 1/3 (-8) + 12*2 +1/3(8)
2(8-8/3)

I don't know what support graphically means, unless it means draw a graph.

check my work.

x = 4 and y = 8

the area = 32