Question

Consider the region \[ R = \left\{ (x, y): x \leq y \leq 9 - \frac{11}{3} x^2, x \geq 0 \right\}. \] The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in \( R \) is:

• A. \( \frac{625}{111} \)
• B. \( \frac{730}{119} \)
• C. \( \frac{567}{121} \)
• D. \( \frac{821}{123} \)

Hint:

When solving optimization problems involving areas, differentiate the area function with respect to the variable (in this case, \( x \)) and solve for the critical points. After finding the critical points, check whether they correspond to a maximum by examining the second derivative or using other methods.

Answer 1

The Correct Option is C

Given the curve: 

\[ y = 9t - \frac{11t^3}{3} \] and the area \( A \) of the rectangle inscribed under this curve is: \[ A = t \left( 9t - \frac{11t^3}{3} \right) = 9t^2 - \frac{11t^4}{3}. \]

Step 1: Differentiate the area function

To find the maximum area, we differentiate the area function with respect to \( t \): \[ \frac{dA}{dt} = 18t - \frac{44t^3}{3}. \]

Step 2: Set the derivative equal to zero

Set \( \frac{dA}{dt} = 0 \) to find critical points: \[ 18t - \frac{44t^3}{3} = 0. \] Multiply through by 3 to eliminate the fraction: \[ 54t - 44t^3 = 0. \] Factor out \( t \): \[ t(54 - 44t^2) = 0. \] Thus, \( t = 0 \) or \( t = \pm \frac{9}{11} \).

Step 3: Determine the maximum point

From the graph and further analysis, we determine that the maximum occurs at \( t = \frac{9}{11} \).

Step 4: Calculate the maximum area

Substituting \( t = \frac{9}{11} \) into the area formula: \[ A = \frac{9}{11} \left( 9 - \frac{11 \cdot 9^3}{3 \cdot 11^3} \right). \] Simplifying: \[ A = \frac{9}{11} \left( 9 - \frac{81}{121} \right) \] \[ A = \frac{9}{11} \times \frac{63}{11} = \frac{567}{121}. \]

Final Answer:

The largest area is \( \frac{567}{121} \).