Question

Let \[ S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x, \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}. \] If the area of the region \( S \) is \( \alpha \sqrt{2} \), then \( \alpha \) is equal to:

• A. \( \frac{17}{2} \)
• B. \( \frac{17}{3} \)
• C. \( \frac{17}{4} \)
• D. \( \frac{17}{5} \)

Hint:

When solving geometry problems involving inequalities, sketch the region to visualize limits of integration.

Answer 1

The Correct Option is B

To find the area of the region \( S \), we compute the integral: \[ \text{Required Area} = \int_0^2 \sqrt{x} \, dx + 1 \cdot 3 \cdot \sqrt{8}. \] Evaluating step by step: \begin{align*} y^2 &= 4x, y^2 = 12 - 2x \Rightarrow x = 2, y = \sqrt{8}
A &= \int_{0}^{2} 2\sqrt{x} \, dx + \frac{1}{2} \times 3 \times \sqrt{8}
&= \left[2 \times \frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{2} + 3\sqrt{2} = \frac{4}{3} \times 2\sqrt{2} + 3\sqrt{2} = \frac{17}{3}\sqrt{2}
\therefore \end{align*} Adding the remaining area: \[ \text{Total Area} = \alpha\sqrt{2} \Rightarrow \alpha = \frac{17}{3} \] Thus, \( \alpha = \frac{17}{3} \).