Consider a rectangle that is inscribed with its base on the x-axis and its upper corners on the parabola y=c−x2, with c> 0. what are the width and height that maximize the area of this rectangle? what is that maximal area?

\frac{4(9c-4)}{27

Step-by-step explanation:

Assuming c positive, we find that this parabola is open down with vertex on the y axis above the origin.

If a rectangle is formed, then it would have two vertices on either side of the y axis of the parabola with remaining two vertices on x axis.

Due to symmetry let us take vertices on x axis as (a,0) and (-a,0)

Corresponding vertices on the parabola would be

Now the rectangle has width = 2b and

length =

So area of the rectangle =

Use derivative test to find a which gives maximum area

Equate I derivative to 0 to get

a =0 or a = 2/3

a cannot be 0

A" is negative for a = 2/3

So maximum when a =2/3

and maximum area

=

if you was gonna solve for a, a is a   < -12

answer: group b.

step-by-step explanation: