Systems of Inequalities: Free Response

Two boundary lines cross at a candidate corner, and a candidate counts only once it also satisfies the constraints that were not used to find it.

On the axes, $4x + y = 24$ meets $y = 0$ at $(6, 0)$, and $2(6) + 5(0) = 12 \le 30$, so that point is feasible. The line $2x + 5y = 30$ meets $x = 0$ at $(0, 6)$, and $4(0) + 6 = 6 \le 24$, so that one is feasible too. The origin satisfies all four constraints, so it is a corner as well.

For the corner on neither axis, solve the two slanted boundaries as a system. From $4x + y = 24$ comes $y = 24 - 4x$, and substituting that into $2x + 5y = 30$ gives

$2x + 5(24 - 4x) = 30 \;\Rightarrow\; -18x = -90 \;\Rightarrow\; x = 5.$

Then $y = 24 - 4(5) = 4$, so the corner is $(5, 4)$. Both slanted constraints hold there with equality, and $x \ge 0$ and $y \ge 0$ hold as well, so it is genuinely in the region.

Two further crossings are not corners at all. The line $2x + 5y = 30$ meets $y = 0$ at $(15, 0)$, where $4(15) + 0 = 60 > 24$, and $4x + y = 24$ meets $x = 0$ at $(0, 24)$, where $2(0) + 5(24) = 120 > 30$. Each fails a constraint it was not built from, so neither belongs to the region.

Evaluate the linear expression $x + y$ at each corner in turn.

$(0,0) \to 0, \quad (6,0) \to 6, \quad (5,4) \to 9, \quad (0,6) \to 6$

The largest of the four is $9$, at $(5, 4)$. On its own that is not yet a statement about the region, which holds infinitely many points while only four were tested.

The corner point principle supplies the missing step, used here in a direction it is not usually used in. What the principle itself says is that wherever a linear expression on a region with corners has a largest value, at least one corner attains it, so the existence of that largest value has to be settled first. On a bounded region it is: slide the line $x + y = c$ upward and the region runs out beneath it, so the slide cannot go on forever, and because the boundaries are solid the last line still meeting the region does meet it. A largest value therefore exists, and the principle then puts it at a corner. The expression $x + y$ is linear, and the region from part A is bounded and has solid boundaries, so its largest value there is among the four just computed. Hence $x + y \le 9$ at every point of that region, corner or not.

Since $9 \le 10$, every point of the part A region already satisfies $x + y \le 10$. Adding that constraint therefore removes no point, and the region cut out by all five constraints is exactly the region cut out by the four.

Note what the principle is doing. It is not being used to find a best plan; it is being used to turn a check at four points into a claim about every point.

Adding two inequalities that point the same way is always legal: if $A \le B$ and $C \le D$ then $A + C \le B + D$. Add the two constraints as they stand, that is with multiplier $1$ on each.

$(2x + 5y) + (4x + y) \le 30 + 24 \;\Rightarrow\; 6x + 6y \le 54$

Dividing by the positive number $6$ keeps the direction.

$x + y \le 9$

Every point of the region satisfies both constraints that were added, so every point satisfies this, corner or not. And $x + y \le 9$ forces $x + y \le 10$, so the fifth constraint holds throughout the region and cuts nothing off it. That is the same conclusion as part B, reached without a single corner.

The bound $9$ is exactly the corner maximum from part B, so this choice of multipliers is the best that these two constraints can do.

The two arguments lean on different things. The combination argument never mentions boundedness at all: it only adds inequalities that every point of the region already satisfies, so it runs word for word whether the region is bounded or not.

The corner argument uses boundedness for one job, to know in advance that a largest value exists. The principle itself asks only that an optimum exist, and then places it at a corner, so a corner argument is still available on an unbounded region whenever the optimum is known to exist and the region has corners to check. An unbounded region can carry a linear objective with a largest value, and can carry one without, so unboundedness on its own settles neither: what it removes is the guarantee, not the possibility.

The combination argument also runs unchanged in three variables or more, where there is no picture to look at, since adding inequalities never mentions how many variables are present.

Look along the $x$-axis, where $4x + y$ grows fastest per unit moved. Setting $y = 0$ and pushing $x$ out, $x + y \le 10$ caps $x$ at $10$ while $2x + 5y \le 30$ would allow $15$, so the binding cap is $10$ and $(10, 0)$ is the far end of what the other constraints permit.

$2(10) + 5(0) = 20 \le 30, \qquad 10 + 0 = 10 \le 10$

So the point clears both of the constraints it has to satisfy, along with $x \ge 0$ and $y \ge 0$, while the constraint under test fails there.

$4(10) + 0 = 40 > 24$

So $(10, 0)$ lies in the region cut out by the other four constraints and outside the region cut out by all five. The two systems have different solution sets, and $4x + y \le 24$ cannot be deleted.

That is what one point establishes. Deleting a constraint safely is a claim about every point of a region at once, so its denial needs only a single witness, and exhibiting one is a complete argument that needs nothing else. An unsuccessful search for positive multipliers proves nothing at all: it rules out the multipliers that were tried, not the existence of others, and it says nothing whatever about the points of the region. The two jobs call for different kinds of evidence, a derivation on one side and a point on the other.

Three boundary lines make three pairs, so take them in turn.

The pair $y = 0$ and $x + y = 12$ gives $x = 12$, the point $(12, 0)$, which satisfies $y \le 5$ and so is feasible. The pair $y = 5$ and $x + y = 12$ gives $x = 7$, the point $(7, 5)$, which satisfies $y \ge 0$ and so is feasible as well. The third pair, $y = 0$ with $y = 5$, is two horizontal lines of the same slope at different heights: they never meet, so that pair produces no candidate at all.

$P(12, 0) = 12, \qquad P(7, 5) = 7 + 10 = 17$

The region therefore has exactly two corners. That count is already worth reading as a signal: a bounded polygon has at least three corners, so two of them says this region is not a polygon.

The move that makes this work is to write $P$ so that the constraints can be substituted straight into it: split $x + 2y$ into a piece that is a constrained left side and a piece left over.

$P = x + 2y = (x + y) + y$

The first piece is bounded by the third constraint and the leftover piece by the second.

$P = (x + y) + y \le 12 + y \le 12 + 5 = 17$

Both steps hold at every point of the region, so $P \le 17$ everywhere: no feasible point, corner or not, beats $17$.

An upper bound on its own promises nothing about being reached, so check the equality case. Equality in the first step needs $x + y = 12$, and equality in the second needs $y = 5$. Both hold together exactly when $y = 5$ and $x = 7$, which is the corner $(7, 5)$, and that point is in the region. So the maximum is $17$, it is attained, and $(7, 5)$ is the only point attaining it.

This argument never asks whether the region is bounded, which is what makes it usable here.

Take any point $(x, y)$ of the region and step one unit to the left, to $(x - 1, y)$. Check it against all three constraints: $y$ is unchanged, so $y \ge 0$ and $y \le 5$ still hold, and $(x - 1) + y = (x + y) - 1 \le 11 \le 12$, so the third holds with room to spare. The new point is feasible.

$P(x - 1, y) = (x - 1) + 2y = P(x, y) - 1$

So every feasible point has a feasible neighbour with a strictly smaller value, which means no feasible point can be the one where $P$ is smallest. There is no minimum.

Repeating the step drives $P$ down without limit, since $P(x - n, y) = P(x, y) - n$ for every whole number $n \ge 0$ and all of those points are feasible. So $P$ is not merely unattained below; it is unbounded below.

What makes this work is that nothing in the system stops $x$ from decreasing. Only $x + y \le 12$ mentions $x$ at all, and it is an upper bound, so it never objects to a smaller $x$. In a situation with a story behind it, the constraint $x \ge 0$ would usually be there without being said; here there is no story, so it is genuinely absent.

The student's rule breaks on this very region. It is unbounded, since part C walks off to the left forever without leaving it, and yet part B proved that $P$ has a maximum of $17$, attained at $(7, 5)$. So "unbounded" cannot imply "no maximum".

What is true runs one way only. A nonempty region that is bounded and includes its boundaries has both a maximum and a minimum for every linear objective, and a corner attains each. Turn that implication around and it fails: a region that is not bounded may still have one of them, or the other, or both.

$Q(12, 0) = 0, \qquad Q(7, 5) = 5$

Test that against $Q = y$ on the same region. Every feasible point has $0 \le y \le 5$, so $Q$ never leaves that band, and both ends are reached, $Q = 5$ at $(7, 5)$ and $Q = 0$ at $(12, 0)$. So an unbounded region can carry an objective with a maximum and a minimum both.

The reason is that unboundedness is a fact about the region alone, while having a maximum is a fact about the region and the objective together. This region runs away in one direction only, straight to the left. An objective whose value falls in that direction, like $P$, keeps its maximum; one that ignores that direction, like $Q$, keeps both; and one that rises in it, such as $-x$, loses its maximum.

Take the sentences one at a time. A wide box holds $2$ trays and a narrow box holds $1$, so $x$ wide boxes and $y$ narrow ones hold $2x + y$ trays, and the school asks for at least $18$ of them.

$2x + y \ge 18$

The lumber caps the total number of boxes, and the rails cap the wide ones alone.

$x + y \le 11, \qquad x \le 4$

Two further constraints belong to the system even though the story never says them out loud: a count of boxes cannot be negative, so $x \ge 0$ and $y \ge 0$. Leaving them out would admit a solution such as $x = -3$, which the situation cannot use.

Counts are also whole numbers, and no inequality here records that. It matters when a plan has to be read off the region, but not for the question of whether any plan exists at all: if no point of the plane satisfies the system, then certainly no whole-number point does.

The tray count $2x + y$ is not itself one of the limits, but it splits into pieces that are.

$2x + y = (x + y) + x$

The lumber caps the first piece at $11$ and the rails cap the second at $4$.

$2x + y = (x + y) + x \le 11 + 4 = 15$

That holds at every point the limits allow, so no build plan whatever reaches more than $15$ trays. Nor is the bound a safe overestimate: at $x = 4$ and $y = 7$ both limits are tight, since $x + y = 11$ and $x = 4$, and that plan holds $2(4) + 7 = 15$ trays. So $15$ is exactly the ceiling.

The order asks for $18$. A point of the region in part A would have to satisfy $2x + y \ge 18$ and $2x + y \le 15$ at the same time, which no point can do. The feasible region is empty: the system has no solution, and under its present limits the club cannot fill the order whatever mix of boxes it tries.

Rebuild the bound with the cap written as a letter. The lumber limit is unchanged, so

$2x + y = (x + y) + x \le 11 + c,$

and the order needs $2x + y \ge 18$, which is impossible unless $11 + c \ge 18$, that is $c \ge 7$. At $c = 6$ the ceiling is $17$, still short, so $7$ is the smallest whole number that can work.

At $c = 7$ the ceiling is exactly $18$ while the order asks for $18$ or more, so every inequality in the chain has to be an equality:

$18 \le 2x + y = (x + y) + x \le 11 + 7 = 18.$

That forces $x + y = 11$ and $x = 7$ together, so $y = 4$. Check the plan: $7 + 4 = 11$ boxes, $7 \le 7$ wide boxes, and $2(7) + 4 = 18$ trays. It works, and nothing else does.

So the region is not a polygon, nor an edge, nor empty: it is a single point. Borrowing three extra rails moves the club from impossible to possible in exactly one way.

A feasible region that is a single point is the tightest outcome short of an empty one. There is exactly one build plan, and every limit is used to the last unit: all $11$ boxes are built, all $7$ rails are used, and the trays come to exactly the $18$ requested and not one more. So the club has no margin anywhere. Lose one rail and the cap drops to $6$, whose ceiling is $17$ trays; lose one board and the box limit drops to $10$, whose ceiling is $10 + 7 = 17$. Either slip makes the order impossible again, and the club would have to renegotiate the order rather than rearrange the mix, because there is no other mix.

Now the second half. Part B never inspects points one at a time. It builds an inequality, $2x + y \le 15$, out of the constraints themselves, and every point satisfying the constraints satisfies it, whether or not anyone ever writes that point down. A drawing works the other way round: it shows the part of the plane that fits on the page at the scale chosen, so "I see no overlap" is a report about a picture rather than a claim about a region. Two half-planes overlapping far off the page look exactly like two that never overlap.

Nothing in the combination step mentions how many variables there are. With three variables the constraints are half-spaces and there is no honest picture at all, yet adding $x + y + z \le 20$ to $z \le 3$ still takes one line.

$(x + y + z) + z \le 20 + 3 \;\Rightarrow\; x + y + 2z \le 23$

A clash between that bound and a fourth constraint would settle emptiness exactly as the clash in part B does here.

The constraints are the two resource limits together with the two counts, neither of which can be negative.

$3x + y \le 21, \quad x + y \le 9, \quad x \ge 0, \quad y \ge 0$

On the axes, $3x + y = 21$ meets $y = 0$ at $(7, 0)$, feasible since $7 + 0 \le 9$, and $x + y = 9$ meets $x = 0$ at $(0, 9)$, feasible since $3(0) + 9 \le 21$. The origin is feasible too. The two remaining axis crossings are not corners: $(9, 0)$ would need $3(9) = 27 \le 21$ and $(0, 21)$ would need $0 + 21 \le 9$, and both fail.

For the corner on neither axis, subtract the trimming equation from the press equation.

$(3x + y) - (x + y) = 21 - 9 \;\Rightarrow\; 2x = 12 \;\Rightarrow\; x = 6$

Then $y = 9 - 6 = 3$, so that corner is $(6, 3)$, and both counts are nonnegative there.

Evaluating $P = kx + y$ at the four corners gives $0$, $7k$, $6k + 3$ and $9$. Two of those carry no $k$ at all, because both $(0, 0)$ and $(0, 9)$ have $x = 0$, which kills the term $kx$.

For each fixed $k$ the corner point principle puts the maximum at a corner, so compare the four expressions from part A. Since $k \ge 0$, the origin's value $0$ never beats the $9$ at $(0, 9)$, so only two comparisons decide the matter.

Against $(0, 9)$:

$6k + 3 > 9 \;\Longleftrightarrow\; 6k > 6 \;\Longleftrightarrow\; k > 1$

Against $(7, 0)$:

$6k + 3 > 7k \;\Longleftrightarrow\; 3 > k$

Both must hold at once, and both hold exactly when $1 < k < 3$.

A check at three sample values. At $k = 2$ the corner values are $0$, $14$, $15$ and $9$, so $(6, 3)$ wins alone; at $k = \frac{1}{2}$ they are $0$, $3.5$, $6$ and $9$, so $(0, 9)$ wins instead; at $k = 4$ they are $0$, $28$, $27$ and $9$, so $(7, 0)$ does.

The strict symbols matter here. At $k = 1$ and at $k = 3$ the winning value is shared, which is what part C is about.

Take the two values in turn, listing the corner values in the order $(0, 0)$, $(7, 0)$, $(6, 3)$, $(0, 9)$.

$k = 1: \quad 0, \quad 7, \quad 9, \quad 9$

So $(6, 3)$ and $(0, 9)$ both give $9$ while $(7, 0)$ gives only $7$. The two winners are the ends of the edge lying on $x + y = 9$, and along that whole edge $P = 1 \cdot x + y = x + y = 9$ identically. The set of maximizing points is therefore not two points but the entire edge between them.

$k = 3: \quad 0, \quad 21, \quad 21, \quad 9$

Now $(7, 0)$ and $(6, 3)$ tie at $21$. They are the ends of the edge lying on $3x + y = 21$, and along it $P = 3x + y = 21$ identically, so again every point of that edge is optimal.

The slopes explain both ties. Solving $kx + y = c$ for $y$ gives $y = c - kx$, so every level line of the objective has slope $-k$. The edge on $x + y = 9$ has slope $-1$, and the edge on $3x + y = 21$ has slope $-3$. A tie along an edge needs the level lines parallel to that edge, that is $-k$ equal to its slope, which is $k = 1$ for the first and $k = 3$ for the second. Parallel is necessary but not sufficient, and the edge must also be one the sliding level line touches last. Take $k = 0$, where the level lines $y = c$ are horizontal and so parallel to the edge on $y = 0$ running from $(0, 0)$ to $(7, 0)$. Every point of that edge does share a value, but the shared value is $0$, the smallest anywhere on the region, while $(0, 9)$ scores $9$ and is the sole maximizer. A level line parallel to an edge the slide meets first ties for the worst value, not the best. Between those two values the level lines are steeper than one of the edges and shallower than the other, which is what pins the maximum to the single corner where those edges meet.

Three of the four corners do win alone somewhere, as parts B and C already show: $(0, 9)$ for $0 \le k < 1$, $(6, 3)$ for $1 < k < 3$, and $(7, 0)$ for $k > 3$.

The origin is a different case, and one comparison disposes of it for every $k$ at once.

$P(0, 0) = k(0) + 0 = 0, \qquad P(0, 9) = k(0) + 9 = 9$

Both points have $x = 0$, so the term $kx$ vanishes at each of them and the parameter cannot influence either value. Whatever $k$ is, $(0, 9)$ scores $9$ and the origin scores $0$, so the origin is beaten and can never be a maximizer, let alone the only one.

That argument is worth more than a sweep. Testing $k = 0, 1, 2, 3$ would show the origin losing four times, which is evidence but not a proof; the comparison above covers every $k$ at once, including the values nobody tried.

In the shop's own terms: printing nothing is never as profitable as printing booklets, because booklets earn a settled positive profit and the poster's price never enters that comparison.