Question

Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is

• A. \( \frac{17}{16} - \log_e 4 \)
• B. \( \frac{33}{8} - \log_e 4 \)
• C. \( \frac{57}{8} - \log_e 4 \)
• D. \( \frac{17}{2} - \log_e 4 \)

Hint:

Always sketch the region when dealing with inequalities. For area bounded between curves, split into simpler subregions and integrate accordingly.

Answer 1

The Correct Option is C

Step 1: List the inequalities defining the region: - \( x>0 \) - \( y>\frac{1}{x} \) - \( 5x - 4y - 1>0 \Rightarrow y<\frac{5x - 1}{4} \) - \( 4x + 4y - 17<0 \Rightarrow y<\frac{17 - 4x}{4} \) So the bounded region lies between: \[ y = \frac{1}{x},\quad y = \min\left(\frac{5x - 1}{4},\ \frac{17 - 4x}{4}\right) \]
Step 2: Find the points of intersection. \( y = \frac{1}{x} \) and \( y = \frac{5x - 1}{4} \): \[ \frac{1}{x} = \frac{5x - 1}{4} \Rightarrow 4 = x(5x - 1) \Rightarrow 5x^2 - x - 4 = 0 \Rightarrow x = 1, -\frac{4}{5} \Rightarrow x = 1 \ (\text{valid}) \] \( y = \frac{1}{x} \) and \( y = \frac{17 - 4x}{4} \): \[ \frac{1}{x} = \frac{17 - 4x}{4} \Rightarrow 4 = x(17 - 4x) \Rightarrow 4x^2 -17x + 4 = 0 \Rightarrow x = \frac{1}{4}, 4 \] So the limits of \( x \) are from \( \frac{1}{4} \) to 1 (for one region) and 1 to 4 (for second).
Step 3: Split region and integrate. Region 1: From \( x = \frac{1}{4} \) to 1, upper curve: \( y = \frac{17 - 4x}{4} \) \[ A_1 = \int_{1/4}^{1} \left[ \frac{17 - 4x}{4} - \frac{1}{x} \right] dx = \int_{1/4}^{1} \left( \frac{17}{4} - x - \frac{1}{x} \right) dx \] \[ = \left[ \frac{17x}{4} - \frac{x^2}{2} - \ln|x| \right]_{1/4}^{1} = \left( \frac{17}{4} - \frac{1}{2} - \ln 1 \right) - \left( \frac{17}{16} - \frac{1}{32} - \ln\left(\frac{1}{4}\right) \right) \] \[ = \left( \frac{15}{4} \right) - \left( \frac{17}{16} - \frac{1}{32} + \ln 4 \right) = \frac{120}{32} - \left( \frac{34 - 1}{32} + \ln 4 \right) = \frac{120 - 33}{32} - \ln 4 = \frac{87}{32} - \ln 4 \] Region 2: From \( x = 1 \) to 4, upper curve: \( y = \frac{5x - 1}{4} \) \[ A_2 = \int_{1}^{4} \left[ \frac{5x - 1}{4} - \frac{1}{x} \right] dx = \int_{1}^{4} \left( \frac{5x}{4} - \frac{1}{4} - \frac{1}{x} \right) dx = \int_{1}^{4} \left( \frac{5x}{4} - \frac{1}{4} - \frac{1}{x} \right) dx \] \[ = \left[ \frac{5x^2}{8} - \frac{x}{4} - \ln|x| \right]_1^4\\ = \left( \frac{80}{8} - \frac{4}{4} - \ln 4 \right) - \left( \frac{5}{8} - \frac{1}{4} - \ln 1 \right) = (10 - 1 - \ln 4) - \left( \frac{5 - 2}{8} \right) = 9 - \ln 4 - \frac{3}{8} = \frac{69}{8} - \ln 4 \]
Step 4: Total area: \[ A = A_1 + A_2 = \left( \frac{87}{32} - \ln 4 \right) + \left( \frac{69}{8} - \ln 4 \right) = \left( \frac{87}{32} + \frac{276}{32} \right) - 2\ln 4 = \frac{363}{32} - 2\ln 4 = \frac{57}{8} - \ln_e 4 \]