a) Let $\mathcal{R}$ be the region consisting of the points $(x,y)$ that satisfy \[-6 \le 3x - 5y \le 6\]and \[-3 \le x + 2y \le
3.\]Plot the region $\mathcal{R},$ and identify the vertices of polygon $\mathcal{R}.$ (b) Find the maximum value of $x + y$ among all points $(x,y)$ in $\mathcal{R}.$
2 answers:
First, we can solve the inequalities as if they were equalities, so we solve the equation 3x - 5y = -6 and x + 2y = 3. Eliminating y, we get x = 3/11. Then y = 15/11, so (x,y) = (3/11, 15/11) is one corner of the region R. The inequalities are symmetric around the origin, so the region R is also symmetric around the region. So R is the rectangle with vertices (-3/11, 15/11), (3/11, 15/11), (-3/11, 15/11), and (-3/11, -15/11).
The maximum value of x + y is then given by the corners of the rectangle R, so the maximum is 3/11 + 15/11 = 18/11.
You might be interested in