Systems of Inequalities: Practice

A corner is where two boundaries cross, so solve the pair. Add the equations to eliminate $y$.

$(x + y) + (x - y) = 6 + 2 \;\Rightarrow\; 2x = 8$

So $x = 4$, and then $y = 6 - x = 2$. The lines meet at $(4, 2)$.

Substitute the coordinates into the objective.

$P = 2(3) + 5(4) = 6 + 20 = 26$

So $P = 26$ at that corner.

By the corner point principle, evaluate $P$ at every vertex and take the largest.

$P(0,0) = 0, \quad P(4,0) = 4, \quad P(0,3) = 12$

The maximum is $12$, at the corner $(0, 3)$.

Evaluate $C$ at each vertex and take the smallest.

$C(2,0) = 6, \quad C(0,5) = 10, \quad C(1,1) = 5$

The minimum is $5$, at the corner $(1, 1)$.

Check all three inequalities.

$2 + 1 = 3 \le 4, \quad 2 \ge 0, \quad 1 \ge 0$

Every inequality holds, so $(2, 1)$ is in the feasible region.

Test both inequalities; failing either one puts the point outside the region.

$2(3) + 3 = 9, \quad 9 \le 8 \text{ is false}$

The first inequality already fails, so $(3, 3)$ is not in the feasible region.

The region is the part of the first quadrant on or above the line $x + y = 2$.

$x + y \ge 2 \text{ has no upper limit on } x + y$

Points can move outward forever toward the upper right, so the region is unbounded.

A point would need $x + y$ to be at most $1$ and at least $4$ at the same time.

$x + y \le 1 \quad\text{and}\quad x + y \ge 4 \text{ cannot both hold}$

No value can be both $\le 1$ and $\ge 4$, so the half-planes never overlap and the region is empty.

Test the known interior point to pick the direction, then choose the symbol from the boundary style.

$0 \le 2(1) = 2 \text{ is true}$

The point satisfies $y \le 2x$, and because the boundary is solid the symbol allows equality, giving $y \le 2x$. A dashed line would instead give $y < 2x$.

The vertical line $x = 2$ fixes the first coordinate and the horizontal line $y = 3$ fixes the second.

$x = 2, \quad y = 3 \;\Rightarrow\; (2, 3)$

The corner is $(2, 3)$.

The line through $(4, 0)$ and $(0, 4)$ is $x + y = 4$. Test the origin to pick the direction.

$0 + 0 = 0 \le 4 \text{ is true}$

The origin satisfies $x + y \le 4$, and the solid boundary allows equality, so the inequality is $x + y \le 4$.

Together the two inequalities demand $x + 2 \le y \le x$.

$x + 2 \le x \;\Rightarrow\; 2 \le 0 \text{ is false}$

No point can have $y$ at least $x + 2$ and at most $x$, so the region is empty and the system has no solution.