Maximizing the area of an inscribed rectangle

A rectangle is inscribed under the parabola $y = 9 - x^2$ , symmetric about the $y$ -axis, so that two corners lie on the parabola at $(x, 9-x^2)$ and $(-x, 9-x^2)$ , and the other two corners lie on the $x$ -axis at $(x,0)$ and $(-x,0)$ , where $0 < x < 3$ .

The width of the rectangle is $2x$ and the height is $9 - x^2$ , so the area is

$A(x) = 2x(9 - x^2), \quad 0 < x < 3.$

Find the value of $x$ that maximizes the area $A(x)$ , and give the corresponding maximum area.

Maximizing the area of an inscribed rectangle graph

Hint

Expand $A(x) = 18x - 2x^3$ , take the derivative $A'(x)$ , set it equal to zero, and solve for $x$ within the domain $0 < x < 3$ . Check that this critical point gives a maximum, then plug it back in to find the area.

Step-by-step solution

Expand the area function as $A(x) = 18x - 2x^3$ . Differentiating gives $A'(x) = 18 - 6x^2$ . Setting $A'(x) = 0$ yields $6x^2 = 18$ , so $x^2 = 3$ and $x = \sqrt{3}$ (taking the positive root since $x > 0$ in the domain).

Since $A''(x) = -12x$ , at $x = \sqrt{3}$ we get $A''(\sqrt{3}) = -12\sqrt{3} < 0$ , confirming a local maximum. Substituting back, $A(\sqrt{3}) = 2\sqrt{3}(9 - 3) = 12\sqrt{3}$ . Thus the maximum area occurs at $x = \sqrt{3}$ and equals $12\sqrt{3}$ .

Maximizing the area of an inscribed rectangle

Related problems