Systems of Inequalities: Practice

The line through $(0, 4)$ and $(2, 0)$ has slope $\frac{0 - 4}{2 - 0} = -2$, so its equation is $y = -2x + 4$, that is $2x + y = 4$. Test the origin for the direction.

$2(0) + 0 = 0 \le 4 \text{ is true}$

The origin satisfies $2x + y \le 4$, and the first quadrant gives $x \ge 0$ and $y \ge 0$, so the system is $x \ge 0$, $y \ge 0$, $2x + y \le 4$.

Two edges lie on the axes, giving $x \ge 0$ and $y \ge 0$. The third edge runs from $(5, 0)$ to $(0, 5)$, the line $x + y = 5$. Test an interior point such as $(1, 1)$.

$1 + 1 = 2 \le 5 \text{ is true}$

The interior satisfies $x + y \le 5$, and every edge is solid, so all three symbols allow equality: $x \ge 0$, $y \ge 0$, $x + y \le 5$.

The corners are $(0, 0)$, $(8, 0)$ (where $2x + y = 16$ meets the $x$-axis), $(0, 10)$ (where $x + y = 10$ meets the $y$-axis), and the crossing of $x + y = 10$ with $2x + y = 16$.

$(2x + y) - (x + y) = 16 - 10 \;\Rightarrow\; x = 6, \; y = 4$

Evaluate $P$: $P(0,0)=0$, $P(8,0)=24$, $P(6,4)=26$, $P(0,10)=20$. The greatest profit is $26$, at $(6, 4)$.

Walk the boundary: the axes and the caps $x \le 5$, $y \le 5$ give $(0,0)$, $(5,0)$, then $x + y = 8$ cuts the corner.

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

The full list is $(0,0)$, $(5,0)$, $(5,3)$, $(3,5)$, $(0,5)$, a pentagon with $5$ corners.

Evaluate $P = x + 3y$ at each corner.

$P(1,4)=13, \quad P(4,1)=7, \quad P(2,2)=8$

The largest value is $13$, so $(1, 4)$ is the maximizer.

The region is the triangle bounded by $y = x$, $y = 4$, and the $y$-axis. Its corners are where those lines cross.

$(0,0),\quad (0,4),\quad (4,4)$

The point $(4, 0)$ fails $y \ge x$, since $0 \ge 4$ is false, so it is not even in the region, let alone a corner.

The corners are $(0,0)$, $(4,0)$ (where $3x + 2y = 12$ meets the $x$-axis), $(0,5)$, and the crossing of $x + y = 5$ with $3x + 2y = 12$.

$3x + 2(5 - x) = 12 \;\Rightarrow\; x = 2, \; y = 3$

Evaluate $P$: $P(0,0)=0$, $P(4,0)=24$, $P(2,3)=27$, $P(0,5)=25$. The maximum is $27$, at $(2, 3)$.

Above the line means $y > -x + 3$. Add $x$ to both sides to write it in standard form.

$y > -x + 3 \;\Rightarrow\; x + y > 3$

The boundary is dashed, so the inequality is strict, giving $x + y > 3$.

The two conditions ask for $x + y$ to be at most $2$ and at least $6$ simultaneously.

$x + y \le 2 \quad\text{and}\quad x + y \ge 6 \text{ is impossible}$

No number is both $\le 2$ and $\ge 6$, so the region is empty.

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

$(3x + 2y) - (x + 2y) = 12 - 8 \;\Rightarrow\; 2x = 4$

So $x = 2$, and then $2y = 8 - 2 = 6$ gives $y = 3$. The corner is $(2, 3)$.

Evaluate $P = 4x + 3y$ at each corner.

$P(0,6)=18, \quad P(3,4)=24, \quad P(5,2)=26$

The greatest profit is $26$, at $(5, 2)$.

Both variables are nonnegative, so $2x + 3y \ge 0$, and the corner $(0, 0)$ reaches that bound.

$P(0, 0) = 2(0) + 3(0) = 0$

The minimum is $0$, at the origin. (The region is unbounded, so there is no maximum, but the minimum exists at the corner.)