Graphing Inequalities and Feasible Regions

Example With $x$ muffins and $y$ cookies, the bake sale has the constraints

The counts cannot be negative, which gives two more constraints, $x \ge 0$ and $y \ge 0$. They come from the context, so they must be written down. The algebra does not supply them on its own.

Example The inequality $y > 2x - 1$ has the boundary line $y = 2x - 1$, through $(0, -1)$ with slope $2$, drawn dashed. The point $(1, 1)$ lies on that line, and $1 > 2 - 1$ is false, so it is not a solution. For $y \ge 2x - 1$ the same line is solid and $(1, 1)$ is a solution.

Example Take $3x - 2y < 6$. Solving for $y$ means dividing by $-2$, which reverses the symbol,

The solutions lie above the dashed line through $(0, -3)$ and $(2, 0)$. A careless reader who sees $<$ in the original and shades below would get exactly the wrong side.

Example Graph $x + 3y \ge 6$. The boundary line passes through $(6, 0)$ and $(0, 2)$, drawn solid. Testing $(0, 0)$ gives $0 \ge 6$, which is false, so the solution is the half-plane that does not contain the origin, the side above the line. For $y < 2x$ the origin lies on the boundary, so test $(1, 0)$ instead. It gives $0 < 2$, true, so shade the side containing $(1, 0)$, below the line.

Example Consider $y \le -x + 4$ and $y > 2x - 2$. The origin satisfies both, since $0 \le 4$ and $0 > -2$, so it is in the solution set of the system. The point $(3, 0)$ satisfies the first, $0 \le 1$, but fails the second, since $0 > 4$ is false. The point $(-2, 7)$ satisfies the second, $7 > -6$, but fails the first, since $7 \le 6$ is false. Each of these two points sits in one shaded half-plane and outside the other, so neither is a solution.

Example For the bake sale, the boundary lines $2x + y = 60$ and $x + y = 40$ meet where subtracting gives $x = 20$, so at $(20, 20)$. Together with $y \ge 0$ the feasible region is the triangle with corners $(20, 20)$, $(30, 0)$ and $(40, 0)$, edges included because every constraint is inclusive.

Only whole numbers of muffins and cookies make sense, so the real choices are the whole number points in the triangle. In the Desmos graphing calculator you can enter each inequality on its own line, for example $y \ge 60 - 2x$, and each one is shaded. The feasible region is the part shaded by every inequality. Enter $x \ge 0$ and $y \ge 0$ too, because zooming the window to the first quadrant does not add them as constraints.