Let the rectangle have its upper right corner at the point (x, 9-x^2) for 0 Find the value of x$ that maximizes the area of the rectangle, and find the corresponding maximum area.

The area of the rectangle is A(x) = (textwidth)(textheight) = 2x(9-x^2) = 18x - 2x^3, defined for 0 0). Since A'(x) > 0 for x √3, this critical point is a maximum. The maximum area is A(√3) = 18√3 - 2(√3)^3 = 18√3 - 6√3 = 12√3.

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Maximizing the area of an inscribed rectangle

A rectangle is inscribed under the parabola $y = 9 - x^2$ so that its base lies on the $x$ -axis and its two upper corners touch the curve, symmetric about the $y$ -axis.

Let the rectangle have its upper right corner at the point $(x, 9-x^2)$ for $0 < x < 3$ , with the rectangle symmetric about the $y$ -axis. The rectangle's width is $2x$ and its height is $9-x^2$ .

Maximizing the area of an inscribed rectangle graph

Find the value of $x$ that maximizes the area of the rectangle, and find the corresponding maximum area.

Hint

Write the area as a function $A(x) = 2x(9 - x^2)$ for $0 < x < 3$ . Take the derivative, set it equal to zero, and solve for $x$ , then check that this critical point gives a maximum.

Step-by-step solution

The area of the rectangle is $A(x) = (\text{width})(\text{height}) = 2x(9-x^2) = 18x - 2x^3$ , defined for $0 < x < 3$ since both the width and height must be positive. This expresses the quantity to be maximized as a function of the single variable $x$ .

Differentiating gives $A'(x) = 18 - 6x^2$ . Setting $A'(x) = 0$ gives $x^2 = 3$ , so $x = \sqrt{3}$ (rejecting the negative root since $x>0$ ). Since $A'(x) > 0$ for $x < \sqrt{3}$ and $A'(x) < 0$ for $x > \sqrt{3}$ , this critical point is a maximum. The maximum area is $A(\sqrt{3}) = 18\sqrt{3} - 2(\sqrt{3})^3 = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}$ .

Maximizing the area of an inscribed rectangle

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