Question

Find the largest possible area of a rectangle with base on x-axis and vertices on curve y = 4 - x^2

Answer 1

The region bounded by the parabola and the x-axis is symmetric, so any inscribed rectangle whose base lies in the x-axis will have its base extend symmetrically around the origin, i.e. if the base has length $2a$, with $a>0$, then its base is the line segment connecting the points $(-a,0)$ and $(a,0)$.

The height of such a rectangle will then by $4-(\pm a)^2=4-a^2$.

The area of such a rectangle is then a function of $a$:

$A(a)=2a(4-a^2)=8a-2a^3$

Differentiating with respect to $a$ gives

$A'(a)=8-6a^2$

which has critical points at

$8-6a^2=0\implies a=\pm\dfrac2{\sqrt3}$

We omit the negative root. Checking the sign of the second derivative at the positive critical point (it's negative) confirms that $a=\dfrac2{\sqrt3}$ is the site of a local maximum.

This means the largest area of this rectangle is

$A\left(\dfrac2{\sqrt3}\right)=\dfrac{32}{3\sqrt3}$