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

To determine the area of the largest rectangle inscribed within the region defined by \(R = \left\{ (x, y): x \leq y \leq 9 - \frac{11}{3} x^2, x \geq 0 \right\}\), we first analyze the region's boundaries.

1. Region boundaries are:
   • The line \(y = x\).
   • The downward-opening parabola \(y = 9 - \frac{11}{3}x^2\), which intersects the y-axis at \(y = 9\).
   • The constraint \(x \geq 0\), confining the region to the right side of the y-axis.
2. The intersection points of \(y = x\) and \(y = 9 - \frac{11}{3}x^2\) are found by setting them equal:
   • \(x = 9 - \frac{11}{3}x^2\).
   • Rearranging yields: \(\frac{11}{3}x^2 + x - 9 = 0\).
3. Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = \frac{11}{3}\), \(b = 1\), and \(c = -9\):
   • The discriminant is \(\Delta = b^2 - 4ac = 1 + \frac{44}{3} \times 9 = 121\).
   • The solutions for \(x\) are \(x = \frac{-1 \pm \sqrt{121}}{\frac{22}{3}} = \frac{-1 \pm 11}{\frac{22}{3}}\).
   • The valid positive x-intersection point is \(x = \frac{10}{\frac{22}{3}} = \frac{30}{11}\).
4. Considering the largest inscribed rectangle within the identified region:
   • The height of the rectangle is given by \(h = x\).
   • The width is determined by the parabola's equation, \(y = 9 - \frac{11}{3} x^2\).
   • The maximum possible height for the rectangle occurs at the intersection point where \(y\) of the parabola equals \(x\), i.e., at \(x = \frac{30}{11}\).
5. The maximum area of the rectangle is calculated as follows:
   • Height \(h = \frac{30}{11}\).
   • Width \(w = 9 - \frac{11}{3}\left(\frac{30}{11}\right)^2\).
   • Substituting \(x = \frac{30}{11}\) to compute the width:
   • \(w = 9 - \frac{11}{3} \left( \frac{900}{121} \right)\)
   • Simplifying the width:
   • \(w = 9 - \frac{9900}{363} = \frac{3267 - 9900}{363} = \frac{567}{121}\).

The area of the largest inscribed rectangle is \(\frac{567}{121}\).