A rectangle in the first quadrant under y = 9 - x^2 has area A = x(9 - x^2). Find x that maximizes A.

A rectangle in the first quadrant under y = 9 - x^2 has area A = x(9 - x^2). Find x that maximizes A.

• A. √3
• B. 3
• C. √6
• D. 0

Answer: (A)

Maximizing the area comes from optimizing A(x) = x(9 − x^2) for x in [0, 3], since the rectangle sits in the first quadrant under the curve y = 9 − x^2. Differentiate: A′(x) = 9 − 3x^2. Set to zero to find critical points: 9 − 3x^2 = 0 ⇒ x^2 = 3 ⇒ x = √3 (in the first quadrant). End points give zero area: A(0) = 0 and A(3) = 0, so the maximum must occur at the critical point. The second derivative A″(x) = −6x is negative at x = √3, confirming a local maximum, which is also the global maximum on the interval because the ends are smaller. Thus, the x value that maximizes the area is √3. The maximum area is A(√3) = √3(9 − 3) = 6√3.

Maximizing the area comes from optimizing A(x) = x(9 − x^2) for x in [0, 3], since the rectangle sits in the first quadrant under the curve y = 9 − x^2.

Differentiate: A′(x) = 9 − 3x^2. Set to zero to find critical points: 9 − 3x^2 = 0 ⇒ x^2 = 3 ⇒ x = √3 (in the first quadrant).

End points give zero area: A(0) = 0 and A(3) = 0, so the maximum must occur at the critical point. The second derivative A″(x) = −6x is negative at x = √3, confirming a local maximum, which is also the global maximum on the interval because the ends are smaller.

Thus, the x value that maximizes the area is √3. The maximum area is A(√3) = √3(9 − 3) = 6√3.