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

Step 1: Define the vertices of the rectangle and its area. 
Consider a rectangle inscribed in the region \( R = \{(x, y) : x \le y \le 9 - \frac{11}{3}x^2, x \ge 0 \} \). 
Let the upper right vertex of the rectangle be \( (x, 9 - \frac{11}{3}x^2) \). 
The lower left vertex is then \( (x,x)\), where \(x \ge 0\). The width of the rectangle is \( x \), and the height is \( 9 - \frac{11}{3}x^2 - x \). 
Therefore, the area of the rectangle, \( A(x) \), is given by: \[ A(x) = x\left(9 - \frac{11}{3}x^2 - x\right) = 9x - x^2 - \frac{11}{3}x^3. \] 
Step 2: Find the critical points by taking the derivative of the area function and setting it to zero. 
To maximize the area, we take the derivative of \( A(x) \) with respect to \( x \) and set it equal to zero: \[ A'(x) = 9 - 2x - 11x^2 = 0. \] \[ 11x^2 + 2x - 9 = 0. \] 
Step 3: Solve the quadratic equation for \( x \). Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(11)(-9)}}{2(11)} = \frac{-2 \pm \sqrt{4 + 396}}{22} = \frac{-2 \pm \sqrt{400}}{22} = \frac{-2 \pm 20}{22}. \] Since \( x \ge 0 \), we take the positive root: \[ x = \frac{-2 + 20}{22} = \frac{18}{22} = \frac{9}{11}. \] Step 4: Calculate the maximum area by plugging the critical point into the area function. Now we substitute \( x = \frac{9}{11} \) into the area function: \[ A\left(\frac{9}{11}\right) = 9\left(\frac{9}{11}\right) - \left(\frac{9}{11}\right)^2 - \frac{11}{3}\left(\frac{9}{11}\right)^3 = \frac{81}{11} - \frac{81}{121} - \frac{11}{3} \cdot \frac{729}{1331} = \frac{81}{11} - \frac{81}{121} - \frac{3}{1} \cdot \frac{243}{121 \cdot 11}=\frac{81}{11} - \frac{81}{121} - \frac{243}{121}. \] \[ A\left(\frac{9}{11}\right) = \frac{81}{11} - \frac{324}{121} = \frac{81 \cdot 11 - 324}{121} = \frac{891 - 324}{121} = \frac{567}{121}. \] Final Answer: The area of the largest rectangle is \( \frac{567}{121} \). The correct answer is (3).