The area of the region {(x,y):0≤x≤94, 0≤y≤1, x≥3y, x+y≥2} is

x+y−2=0P(32,12);Q(2,0);R(94,0);S(94,34)Area =12||322,120|+|294,00|+|9494034|+|94323412||=12|(0−1)+(0−0)+(2716−0)+(98−98)|=1132

The area of the region {(x,y):0≤x≤94, 0≤y≤1, x≥3y, x+y≥2} is

x+y−2=0P(32,12);Q(2,0);R(94,0);S(94,34)Area =12||322,120|+|294,00|+|9494034|+|94323412||=12|(0−1)+(0−0)+(2716−0)+(98−98)|=1132

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The area of the region ${(x, y) : 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3 y, x + y \geq 2}$ is

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$\frac{37}{96}$

$\frac{35}{96}$

$\frac{11}{32}$

$\frac{13}{32}$

answer is A.

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Detailed Solution

$\begin{array}{l} x + y - 2 = 0 \\ P (\frac{3}{2}, \frac{1}{2}); Q (2, 0); R (\frac{9}{4}, 0); S (\frac{9}{4}, \frac{3}{4}) \\ A r e a = \frac{1}{2} | | \begin{matrix} \frac{3}{2} \\ 2 \end{matrix}, \begin{matrix} \frac{1}{2} \\ 0 \end{matrix} | + | \begin{matrix} 2 \\ \frac{9}{4} \end{matrix}, \begin{matrix} 0 \\ 0 \end{matrix} | + | \begin{matrix} \frac{9}{4} \\ \frac{9}{4} \end{matrix} \begin{matrix} 0 \\ \frac{3}{4} \end{matrix} | + | \begin{matrix} \frac{9}{4} \\ \frac{3}{2} \end{matrix} \begin{matrix} \frac{3}{4} \\ \frac{1}{2} \end{matrix} | | \\ = \frac{1}{2} | (0 - 1) + (0 - 0) + (\frac{27}{16} - 0) + (\frac{9}{8} - \frac{9}{8}) | = \frac{11}{32} \end{array}$