91Ó°ÊÓ

The first inequality is \(|\mathrm{x}|+|\mathrm{y}| \leq 3\). As we know, the graph of \(|\mathrm{x}|+|\mathrm{y}| = k\) forms a square with side length \(2k\) centered at the origin and oriented with its diagonal parallel to the axes. So we have a square of side length \(6\) and vertices \((\pm3, \pm3)\).

The second inequality is \(\mathrm{xy} \geq 2\). This inequality is positive when both x and y have the same sign and when their product exceeds 2. This has the effect of creating a region in the first and third quadrants where the inequality is satisfied.

Now we need to find the intersection points between these two inequalities. Let's solve for the intersection points, which is equivalent to finding the points where the boundaries intersect, that is, \(|\mathrm{x}|+|\mathrm{y}|=3\) and \(\mathrm{xy} =2\). We can break the first equation into two cases: Case 1: \(x \geq 0\) and \(y \geq 0\) i) \(x + y = 3\) and \(xy = 2\). Solving for y in the first equation, we get \(y = 3 - x\). Substituting it in the second equation, we get \(x(3 - x) = 2\) or \(x^2-3x+2=0\). Factoring the equation gives \((x - 1)(x - 2) = 0\). So we have two intersection points in the first quadrant: \((1, 2)\) and \((2, 1)\). Case 2: \(x \leq 0\) and \(y \leq 0\) ii) \(-x - y = 3\) and \(xy = 2\). Similar to the previous case, we get \(y = -3 + x\) and inserting this into the second equation gives \(-x^2+3x-2=0\). Factoring the equation gives \((-x + 1)(-x + 2) = 0\). So we have two intersection points in the third quadrant: \((-1, -2)\) and \((-2, -1)\).

Now it's time to compute the area enclosed by these two inequalities. We can split the area into two symmetrical parts - the one in the first quadrant and the one in the third quadrant. Due to symmetry, the area of each part will be the same, so we can compute the area in the first quadrant and then multiply by 2 to get the total area. In the first quadrant, we have the intersection points \((1, 2)\) and \((2, 1)\). The area enclosed by the inequalities consists of two right triangles formed by the vertices \(O(0,0)\), \((1,2)\), and \((2,1)\), where O is the origin. The area of each triangle can be found using the formula \(\frac{1}{2} \cdot base \cdot height\). Since the two triangles have the same base and height but different orientations, their areas are equal. Triangle 1 area: \(\frac{1}{2}(1)(2) = 1\) Triangle 2 area: \(\frac{1}{2}(2)(1) = 1\) Now, the total area enclosed in the first quadrant is \(1+1=2\). Since the area in the third quadrant is symmetrical, the total area enclosed by the given inequalities is \(2 + 2 = 4\).