Systems of Inequalities: Additional Practice (Set 2)

The region is unbounded, but its lower boundary $x + y = 4$ has the two corners $(4, 0)$ and $(0, 4)$, and the minimum of a linear objective sits at a corner.

$C(4, 0) = 12, \quad C(0, 4) = 16$

The minimum is $12$, at $(4, 0)$. There is no maximum, since the region runs outward forever, but the minimum still exists at a vertex.

The corners are $(0,0)$, $(7,0)$ (where $2x + y = 14$ meets the $x$-axis), $(0,9)$, and the crossing of $2x + y = 14$ with $x + y = 9$.

$(2x + y) - (x + y) = 14 - 9 \;\Rightarrow\; x = 5, \; y = 4$

Evaluate $P$: $P(0,0)=0$, $P(7,0)=35$, $P(5,4)=41$, $P(0,9)=36$. The greatest profit is $41$ dollars, at $(5, 4)$.

The first quadrant extends forever, so $x + y$ can be made as large as you like.

$\text{Taking } (x, y) = (n, n) \text{ gives } P = 2n \to \infty$

No finite value is the largest, so $P$ has no maximum. (It does have a minimum of $0$ at the origin, but that is a different question.)

A region is empty when the constraints contradict each other. Compare the sum $x + y$ in each choice.

$x + y \le 1 \quad\text{and}\quad x + y \ge 5 \text{ cannot both hold}$

No value is both $\le 1$ and $\ge 5$, so the system with $x + y \le 1$ and $x + y \ge 5$ is empty. Each other system has points that satisfy both of its inequalities.

Evaluate $P$ at all four corners.

$P(0,0)=0, \; P(4,0)=4, \; P(4,3)=10, \; P(0,3)=6$

The maximum is $10$, at $(4, 3)$.

Subtract the second equation from the first to eliminate $y$.

$(4x + y) - (x + y) = 9 - 3 \;\Rightarrow\; 3x = 6$

So $x = 2$, and then $y = 3 - x = 1$. The corner is $(2, 1)$.

The region is unbounded above, so check the corners of its lower boundary: $(6, 0)$, $(0, 4)$, and the crossing of $x + y = 4$ with $x + 3y = 6$.

$(x + 3y) - (x + y) = 6 - 4 \;\Rightarrow\; 2y = 2 \;\Rightarrow\; (3, 1)$

Evaluate $C$: $C(6,0)=12$, $C(3,1)=9$, $C(0,4)=12$. The minimum is $9$, at $(3, 1)$.

The region is unbounded above, so check the corners of its lower boundary: $(0, 8)$, $(5, 0)$, and the crossing of $2x + y = 8$ with $x + y = 5$.

$(2x + y) - (x + y) = 8 - 5 \;\Rightarrow\; x = 3, \; y = 2$

Evaluate $C$: $C(0,8)=40$, $C(3,2)=22$, $C(5,0)=20$. The minimum is $20$, at $(5, 0)$.

Check every inequality at $(2, 2)$.

$x + 2y = 6 \le 6, \quad 2x + y = 6 \le 6, \quad 2 \ge 0, \quad 2 \ge 0$

All four hold (the first two with equality, so $(2, 2)$ is actually the corner where those boundaries cross). The point is in the region.

Points to the left of the line have a smaller $x$-coordinate, and the solid line includes its own points.

$x \le 4$

The symbol $\le$ allows equality, matching the solid boundary; a dashed line would give $x < 4$.

Two edges lie on the axes ($x \ge 0$, $y \ge 0$). The third runs from $(6, 0)$ to $(0, 4)$: its equation is $2x + 3y = 12$, since both points satisfy it. Test the interior point $(1, 1)$.

$2(1) + 3(1) = 5 \le 12 \text{ is true}$

The interior satisfies $2x + 3y \le 12$, so the system is $x \ge 0$, $y \ge 0$, $2x + 3y \le 12$.

Check each inequality at $(1, 3)$.

$1 + 3 = 4 \le 5, \quad 3 \le 2(1) + 1 = 3, \quad 1 \ge 0$

All three hold (the second with equality, so the point sits on that boundary). The point is in the region.