Question

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?

Answer 1

Answer:

\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

$(b, c-a^2) and (b,c-a^2)$

Now the rectangle has width = 2b and

length = $c-a^2$

So area of the rectangle =

$A(a) =2a(c-a^2)\\A(a) = 2ac-a^3$

Use derivative test to find a which gives maximum area

$A'(a) = 2a-3a^2\\A"(a) = 2-6a$

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

= $2(\frac{2}{3} )(c-\frac{4}{9})\\ =\frac{4(9c-4)}{27}$