The problem asks for the area of the region

defined by the following inequalities:

\({(x,y):0\leq x\leq \frac{9}{4},0\leq y\leq 1,x\geq 3y,x+y\geq 2}\)

First, let's rewrite the inequalities in terms of \(y\):

\(0\leq x\leq \frac{9}{4}\)

\(0\leq y\leq 1\)

\(x\geq 3y\text{ }⟹\text{ }y\leq \frac{x}{3}\)

\(x+y\geq 2\text{ }⟹\text{ }y\geq 2−x\)

Combining these, for a given \(x\), the range of \(y\) is:

\(\max ⁡(0,2−x)\leq y\leq \min ⁡(1,\frac{x}{3})\)

For the region to be non-empty,

we must have \(\max ⁡(0,2−x)\leq \min ⁡(1,\frac{x}{3})\).

Let's find the critical points by examining

the intersections of the boundary lines.

Intersection of \(y=2−x\) and \(y=\frac{x}{3}\):

\(\begin{matrix}2−x & =\frac{x}{3} \\ 6−3x & =x \\ 6 & =4x \\ x & =\frac{6}{4}=\frac{3}{2}\end{matrix}\)

At \(x=\frac{3}{2}\), \(y=2−\frac{3}{2}=\frac{1}{2}\). This point is \({P}_{1}(\frac{3}{2},\frac{1}{2})\).

This point satisfies \(0\leq \frac{3}{2}\leq \frac{9}{4}\)

(since \(1.5\leq 2.25\)) and \(0\leq \frac{1}{2}\leq 1\).

The condition \(y\geq 2−x\) implies that for \(x<\frac{3}{2}\), \(2−x>\frac{x}{3}\),

so the lower bound \(2−x\) would be greater than

the upper bound \(\frac{x}{3}\), meaning no region

exists for \(x<\frac{3}{2}\). Thus, the region starts at \(x=\frac{3}{2}\).

The upper limit for \(x\) is \(\frac{9}{4}\).

The upper bound for \(y\) is \(y=\frac{x}{3}\). Since \(x\leq \frac{9}{4}=2.25\),

then \(\frac{x}{3}\leq \frac{9\mathrm{/}4}{3}=\frac{3}{4}\). This is always less than or equal to \(1\),

so \({y}_{upper}=\frac{x}{3}\) for the entire \(x\)-range of the region.

The lower bound for \(y\) is \(y=\max ⁡(0,2−x)\).

If \(2−x\geq 0\text{ }⟹\text{ }x\leq 2\), then \({y}_{lower}=2−x\).

If \(2−x<0\text{ }⟹\text{ }x>2\), then \({y}_{lower}=0\).

So we need to split the integral into

two parts based on the lower bound for \(y\).

The relevant \(x\)-interval is \([\frac{3}{2},\frac{9}{4}]\).

The critical \(x\)-value where

the lower bound changes is \(x=2\).

Define the integration intervals

Interval 1: \(\frac{3}{2}\leq x\leq 2\)

In this interval, \({y}_{upper}=\frac{x}{3}\) and \({y}_{lower}=2−x\).

Interval 2: \(2

In this interval, \({y}_{upper}=\frac{x}{3}\) and \({y}_{lower}=0\).

Calculate the area for Interval 1

The area \({A}_{1}\) is given by:

\(\begin{matrix}{A}_{1} & ={\int }_{\frac{3}{2}}^{2}(\frac{x}{3}−(2−x))dx \\ & ={\int }_{\frac{3}{2}}^{2}(\frac{x}{3}−2+x)dx \\ & ={\int }_{\frac{3}{2}}^{2}(\frac{4x}{3}−2)dx \\ & ={[\frac{4}{3}⋅\frac{{x}^{2}}{2}−2x]}_{\frac{3}{2}}^{2} \\ & ={[\frac{2{x}^{2}}{3}−2x]}_{\frac{3}{2}}^{2} \\ & =(\frac{2(2{)}^{2}}{3}−2(2))−(\frac{2{(\frac{3}{2})}^{2}}{3}−2(\frac{3}{2})) \\ & =(\frac{8}{3}−4)−(\frac{2⋅\frac{9}{4}}{3}−3) \\ & =(\frac{8}{3}−\frac{12}{3})−(\frac{9}{6}−3) \\ & =−\frac{4}{3}−(\frac{3}{2}−\frac{6}{2}) \\ & =−\frac{4}{3}−(−\frac{3}{2}) \\ & =−\frac{4}{3}+\frac{3}{2} \\ & =\frac{−8+9}{6}=\frac{1}{6}\end{matrix}\)

Calculate the area for Interval 2

The area \({A}_{2}\) is given by:

\(\begin{matrix}{A}_{2} & ={\int }_{2}^{\frac{9}{4}}(\frac{x}{3}−0)dx \\ & ={\int }_{2}^{\frac{9}{4}}\frac{x}{3}dx \\ & ={[\frac{{x}^{2}}{6}]}_{2}^{\frac{9}{4}} \\ & =\frac{{(\frac{9}{4})}^{2}}{6}−\frac{(2{)}^{2}}{6} \\ & =\frac{\frac{81}{16}}{6}−\frac{4}{6} \\ & =\frac{81}{96}−\frac{64}{96} \\ & =\frac{17}{96}\end{matrix}\)

Calculate the total area

The total area \(A\) is the sum of \({A}_{1}\) and \({A}_{2}\):

\(\begin{matrix}A & ={A}_{1}+{A}_{2} \\ & =\frac{1}{6}+\frac{17}{96} \\ & =\frac{16}{96}+\frac{17}{96} \\ & =\frac{33}{96}\end{matrix}\)

Simplifying the fraction by dividing

the numerator and denominator by 3:

\(A=\frac{33\div 3}{96\div 3}=\frac{11}{32}\)

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

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