Systems of Inequalities: Additional Practice (Set 1)

The corners are $(0,0)$, $(7,0)$ (where $x + y = 7$ meets the $x$-axis), $(0,5)$ (where $x + 2y = 10$ meets the $y$-axis), and the crossing of the two slanted lines.

$(x + 2y) - (x + y) = 10 - 7 \;\Rightarrow\; y = 3, \; x = 4$

Evaluate $P$: $P(0,0)=0$, $P(7,0)=21$, $P(4,3)=27$, $P(0,5)=25$. The maximum is $27$, at $(4, 3)$.

The line through $(0, 6)$ and $(3, 0)$ has slope $\frac{0 - 6}{3 - 0} = -2$, so it is $y = -2x + 6$, that is $2x + y = 6$. Test the origin.

$2(0) + 0 = 0 < 6 \text{ is true}$

The origin satisfies $2x + y < 6$, and because the boundary is dashed the inequality is strict. With the first quadrant this gives $x \ge 0$, $y \ge 0$, $2x + y < 6$.

Trace the boundary: $(0,0)$, $(6,0)$ along the $x$-axis, up $x = 6$ to where $x + y = 8$, then across to where $y = 5$, then down the $y$-axis.

$(6, 2) \text{ where } x = 6, \quad (3, 5) \text{ where } y = 5$

The corners are $(0,0)$, $(6,0)$, $(6,2)$, $(3,5)$, $(0,5)$: a pentagon with $5$ corners.

Evaluate $P$ at every corner.

$P(0,0)=0, \; P(5,0)=10, \; P(3,4)=10, \; P(0,6)=6$

The maximum value is $10$. It happens to be reached at two corners, $(5, 0)$ and $(3, 4)$, which occurs when the objective line is parallel to the edge joining them; the value is still $10$.

The region is unbounded, but $x + y$ is smallest when both variables are as small as allowed, at the corner $(2, 1)$.

$P(2, 1) = 2 + 1 = 3$

The minimum is $3$. The region has no upper bound, so $P$ has no maximum, but the minimum exists at the corner.

Add the two inequalities. The left sides cancel to $0$.

$(x - y) + (y - x) \ge 3 + 3 \;\Rightarrow\; 0 \ge 6 \text{ is false}$

Assuming a common solution leads to the false statement $0 \ge 6$, so no point satisfies both and the region is empty.

Evaluate $P$ at each corner.

$P(0,0)=0, \; P(3,0)=21, \; P(2,2)=26, \; P(0,3)=18$

The maximum is $26$, at $(2, 2)$.

The region is the part of the first quadrant on or above the line $x + y = 3$.

$x + y \ge 3 \text{ places no ceiling on } x + y$

Points extend outward without limit, so the region is unbounded.

Evaluate $P = x + y$ at each corner.

$P(1,1)=2, \quad P(2,5)=7, \quad P(4,2)=6$

The maximum is $7$, at $(2, 5)$.

Substitute $y = 2x$ into the second equation.

$x + 2x = 6 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

Then $y = 2x = 4$, so the corner is $(2, 4)$.

The region is unbounded, so check the corners of its lower boundary $x + y = 8$, namely $(8, 0)$ and $(0, 8)$.

$C(8, 0) = 24, \quad C(0, 8) = 16$

The minimum is $16$, at $(0, 8)$. Because $y$ is cheaper per unit than $x$ in the objective, pushing everything onto $y$ is best.

Trace the boundary: $(0,0)$, $(4,0)$ along the $x$-axis, up $x = 4$ to where $x + y = 6$, then across to where $y = 4$, then down the $y$-axis.

$(4, 2) \text{ where } x = 4, \quad (2, 4) \text{ where } y = 4$

The corners are $(0,0)$, $(4,0)$, $(4,2)$, $(2,4)$, $(0,4)$: a pentagon with $5$ corners.