Example 4: $Z = -50x + 20y$ on an unbounded region: no minimum

Basics you need

Part (i): the minimum of $Z$

1. Draw the line $2x - y = -5$ and choose its side

   Put $y = 0$⁠: $2x = -5$⁠, so $x = -\frac52$⁠. Put $x = 0$⁠: $-y = -5$⁠, so $y = 5$⁠. The line joins $\left(-\frac52, 0\right)$ and $(0, 5)$⁠; it also passes through $(1, 7)$⁠.

   Test $(0, 0)$ in (2): $0 \ge -5$ is true, so (2) keeps the side of the line that contains the origin. In the first quadrant this is the side below the line: (2) says $y \le 2x + 5$⁠.

   The $x$-axis point is to the left of the origin, outside the first quadrant, but it still fixes the line. The test point needs care with the minus signs: $2(0) - 0 = 0$⁠, and $0$ is greater than $-5$⁠.
   Basics: where a line meets the axes · Basics: which side

2. Draw the line $3x + y = 3$ and choose its side

   Its axis points are $(1, 0)$ and $(0, 3)$⁠. Test $(0, 0)$ in (3): $0 \ge 3$ is false, so (3) keeps the side away from the origin.

   "At least 3" fails at the origin, so the kept side is the far one.
   Basics: one side of a line

3. Draw the line $2x - 3y = 12$ and choose its side

   Put $y = 0$⁠: $x = 6$⁠. Put $x = 0$⁠: $-3y = 12$⁠, so $y = -4$⁠. The line joins $(6, 0)$ and $(0, -4)$⁠; it also passes through $(9, 2)$⁠.

   Test $(0, 0)$ in (4): $0 \le 12$ is true, so (4) keeps the origin's side, which is the side above the line: (4) says $y \ge \frac{2x - 12}{3}$⁠.

   Again one axis point, $(0, -4)$⁠, is outside the first quadrant. Rewriting (4) as $3y \ge 2x - 12$ (adding $3y - 12$ to both sides) shows directly that the kept side is above the line.
   Basics: one side of a line

4. Shade the feasible region: it is unbounded

   In the first quadrant, the region lies below $2x - y = -5$⁠, beyond $3x + y = 3$⁠, and above $2x - 3y = 12$⁠. It is shaded below; the darker part and the dashed line are used in Part (i), Step 7.

   The region is unbounded: every point $(t, t)$ with $t \ge 1$ is in it ($t \ge -5$⁠, $4t \ge 3$⁠, $-t \le 12$⁠), and these points go as far from O as we like, so no circle can enclose the region.

   Between the two upward-sloping lines $y = 2x + 5$ and $y = \frac{2x - 12}{3}$ there is a strip that keeps widening to the right, because the first line is steeper. Nothing closes it off, which is what "unbounded" means.
   Basics: where the shadings overlap · Basics: bounded or not

5. Find the corner points

   $(0, 5)$⁠: line (2) meets the $y$-axis. $(0, 3)$⁠: line (3) meets the $y$-axis. $(1, 0)$⁠: line (3) meets the $x$-axis. $(6, 0)$⁠: line (4) meets the $x$-axis. Each obeys every constraint; for example $(6, 0)$⁠: $12 \ge -5$⁠, $18 \ge 3$⁠, $12 \le 12$⁠.

   No other crossing point is in the region. Lines (2) and (3) meet at $\left(-\frac25, \frac{21}{5}\right)$⁠, lines (2) and (4) at $\left(-\frac{27}{4}, -\frac{17}{2}\right)$⁠, lines (3) and (4) at $\left(\frac{21}{11}, -\frac{30}{11}\right)$⁠: each has a negative coordinate. $\left(-\frac52, 0\right)$ and $(0, -4)$ are outside the first quadrant, and O breaks (3).

   An unbounded region still has corners; it just also has edges that run off for ever (here along lines (2) and (4)). Checking all the crossing points makes sure the list of corners is complete.
   Basics: a corner solves two equations

6. Work out $Z$ at each corner

   Corner point	$Z = -50x + 20y$
   $(0, 5)$	$0 + 100 = 100$ ← largest
   $(0, 3)$	$0 + 60 = 60$
   $(1, 0)$	$-50 + 0 = -50$
   $(6, 0)$	$-300 + 0 = -300$ ← smallest

   So $m = -300$⁠, at $(6, 0)$⁠.

   If the region were bounded, Theorem 2 would make $-300$ the minimum at once. It is unbounded, so $-300$ may or may not be the minimum, and the next step decides.
   Lesson: the corner point method

7. The open half-plane test for $m = -300$

   Following step 3(ii) of the Corner Point Method, we look at the open half-plane

   $$\begin{aligned} -50x + 20y &< -300, \\ \text{i.e.}\quad -5x + 2y &< -30 \end{aligned}$$

   (dividing by 10). Its boundary $-5x + 2y = -30$ passes through $(6, 0)$ and $(10, 10)$⁠; it is drawn dashed because the line itself is not included.

   The feasible point $(9, 2)$ is in this half-plane: $-5(9) + 2(2) = -41 < -30$⁠, and indeed $Z(9, 2) = -450 + 40 = -410 < -300$⁠. So the half-plane has points in common with the feasible region (the darker part of the graph), and $Z$ has no minimum value.

   The half-plane $Z < -300$ contains exactly the points where $Z$ is smaller than the best corner value. A feasible point there shows $-300$ is not the smallest value. In fact $Z$ has no smallest value at all: along the edge $2x - 3y = 12$⁠, at $(6 + 3t, 2t)$⁠, $Z = -300 - 110t$⁠, which falls below any number as $t$ grows. The book prints the two tests of step 3(ii) as items "4. (a)" and "4. (b)" just below step 3(ii); the test used here is the one for the minimum, (b), and the book's text of Example 4 points to it as step 3(ii) of the Corner Point Method.
   Lesson: the half-plane check, Example 4 · Basics: lines of equal Z

Part (ii): the book's follow-up, and why the half-plane is checked

1. Why we check $-50x + 20y > 100$

   The largest corner value is $M = 100$⁠, at $(0, 5)$⁠. Because the region is unbounded, this is not automatically the maximum: $Z$ might be larger at some point far out in the region. The test of the Corner Point Method (printed as item 4(a), just below step 3(ii)) says: $M$ is the maximum if the open half-plane $-50x + 20y > 100$ has no point in common with the feasible region; otherwise $Z$ has no maximum value.

   This answers the book's "(Why?)". The open half-plane $Z > 100$ is exactly the set of points where $Z$ beats the best corner. If a feasible point lay in it, 100 would not be the maximum. If none does, then $Z \le 100$ at every feasible point, and $Z = 100$ is actually reached at $(0, 5)$⁠, so 100 is the maximum.
   Lesson: unbounded regions

2. Draw the half-plane

   Dividing by 10, the half-plane is $-5x + 2y > 10$⁠. Its boundary $-5x + 2y = 10$ passes through $(-2, 0)$ and $(0, 5)$⁠, and the half-plane is the side above and to the left of it (the origin gives $0 > 10$⁠, false).

   The dashed line is steeper (slope $\frac52$⁠) than the edge $2x - y = -5$ (slope 2), and both pass through $(0, 5)$⁠. So to the right of the $y$-axis the dashed line runs above the region's upper edge.
   Basics: one side of a line

3. Show that there is no common point

   Every feasible point satisfies (2), that is $y \le 2x + 5$⁠, and $x \ge 0$⁠. Hence

   $$\begin{aligned} -5x + 2y &\le -5x + 2(2x + 5) \\ &= 10 - x \\ &\le 10. \end{aligned}$$

   So $-5x + 2y > 10$ never holds on the feasible region: the open half-plane has no point in common with it.

   Multiplying $y \le 2x + 5$ by the positive number 2 keeps the direction. The graph suggests the answer; this one line of algebra proves it for the whole infinite region.
   Basics: where the shadings overlap

4. Conclude

   Yes: $Z = -50x + 20y$ has the maximum value 100 at $(0, 5)$. Multiplying the last result by 10, $Z \le 100 - 10x \le 100$ at every feasible point, with equality only when $x = 0$ and $y = 5$⁠.

   So on this unbounded region $Z$ has a maximum but no minimum. Unboundedness only makes an optimum possible to lose; the half-plane test decides each case separately.
   Lesson: Example 4, step by step

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