Introduction to Optimization: Free Response

Two vertices come straight from the axes. On the $x$-axis ($y=0$), the binding limit is $2x + y = 8$, giving $x = 4$, so $(4,0)$. On the $y$-axis ($x=0$), the binding limit is $x + 3y = 9$, giving $y = 3$, so $(0,3)$.

The fourth vertex is where the two slanted lines meet. Solve $x + 3y = 9$ for $x$ and substitute into $2x + y = 8$:

$2(9 - 3y) + y = 8 \;\Rightarrow\; 18 - 6y + y = 8 \;\Rightarrow\; -5y = -10 \;\Rightarrow\; y = 2,$

so $x = 9 - 3(2) = 3$: the vertex $(3,2)$. Together with the origin, the four vertices are $(0,0)$, $(4,0)$, $(0,3)$, and $(3,2)$.

Substitute each vertex into the objective.

$P(0,0) = 0, \qquad P(4,0) = 3(4) = 12, \qquad P(0,3) = 4(3) = 12, \qquad P(3,2) = 3(3) + 4(2) = 9 + 8 = 17.$

The largest of the four values is $17$, reached at $(3,2)$, the vertex that is not on either axis.

The corner-point principle says that over a bounded region like this one the maximum is reached at a vertex, but it makes no promise about WHICH vertex; every one of them is a candidate until it has been checked. A region bounded by two slanted lines together with the axes generally has a vertex where the two slanted lines cross each other, and that point is just as much a vertex as the two on the axes.

$\text{axis vertices alone} \subsetneq \text{every vertex of the region}$

Stopping at the axis intercepts silently drops a candidate from the list, so the largest of the remaining values is only the best of an incomplete search, not necessarily the true maximum. In this problem that dropped vertex is exactly the one that wins.

Solve $x + 2y = 10$ for $x$: $x = 10 - 2y$. Substitute into $3x + y = 15$:

$3(10 - 2y) + y = 15 \;\Rightarrow\; 30 - 6y + y = 15 \;\Rightarrow\; -5y = -15 \;\Rightarrow\; y = 3,$

so $x = 10 - 2(3) = 4$: the vertex $(4,3)$.

Take any point strictly inside the feasible region: it lies on no boundary, so a small enough move in any direction stays inside the region. In particular, move a small step in the direction that increases $c$, the direction the level lines slide toward larger values. That step keeps every constraint satisfied, because the point started with room to spare on every one of them, and it lands on a level line with $P$ larger than before.

$\text{interior point} \;\Rightarrow\; \text{a nearby feasible point with strictly larger } P \text{ always exists}$

So from any interior point there is always a nearby feasible point with a strictly larger value of $P$. A point that can always be beaten cannot be the maximum. This rules out every interior point at once, which is why, on a bounded region like this one, the maximum has to sit on the boundary and ultimately at a vertex of it.

First check that $(2,3)$ really is strictly interior: $x + 2y = 2 + 6 = 8$, strictly less than $10$, and $3x + y = 6 + 3 = 9$, strictly less than $15$, with $x = 2 > 0$ and $y = 3 > 0$ as well. Every one of the four constraints has room to spare, so the point touches no boundary.

$P(2,3) = 2(2) + 5(3) = 4 + 15 = 19.$

That $19$ looks like a respectable value, but part B's argument applies to it exactly as it does to any interior point: a small step from $(2,3)$ toward larger $P$ stays feasible and lands on a strictly larger value. So $19$ is provably not the maximum, regardless of how it compares to any particular vertex; only checking the actual vertices can settle the question.

Each vertex is where $x + y = 3$ meets an axis. On the $x$-axis ($y=0$): $x = 3$, giving $(3,0)$. On the $y$-axis ($x=0$): $y = 3$, giving $(0,3)$. No other pair of boundaries meets in the first quadrant, since $x \ge 0$ and $y \ge 0$ do not cross each other except at the origin, and the origin fails $x + y \ge 3$.

$P(3,0) = 3 + 0 = 3, \qquad P(0,3) = 0 + 3 = 3.$

Both vertices give the objective the same value.

The claim says the largest vertex value is always the true maximum. Pick any feasible point that is not one of the two vertices and see whether it beats them. Take $(4,4)$: check it is feasible first, $4 + 4 = 8 \ge 3$ with $x, y \ge 0$, so yes.

$P(4,4) = 4 + 4 = 8.$

The largest vertex value from part A was $3$, and $8 > 3$. A feasible point has beaten every vertex value, which is exactly what the claim said could never happen. One such point is enough to refute a claim made about ALL feasible regions, so the claim is false.

In part B, moving away from the vertices toward larger $x$ and $y$ never left the region: it kept growing, further than any bound. That is exactly what an unbounded region means, and it is why no vertex value could be trusted as the largest possible: whatever value a vertex gives, a feasible point farther out gives more.

$\text{unbounded region} \;\Rightarrow\; \text{no vertex value can be trusted as the largest possible}$

The corner-point principle, as proved in this lesson, guarantees both a maximum and a minimum only for a BOUNDED feasible region, one that fits inside some large enough box. Here the region stops growing toward the origin, bounded below by $x+y \ge 3$, so a minimum exists, but it never stops growing away from the origin, so no maximum exists.

Wax used is $3$ oz per vanilla candle and $2$ oz per lavender candle, and the total cannot exceed the $18$ oz available:

$3x + 2y \le 18.$

Curing time used is $1$ hour per vanilla candle and $2$ hours per lavender candle, capped at $10$ hours:

$x + 2y \le 10.$

Both counts are nonnegative, $x \ge 0$ and $y \ge 0$, and the profit to maximize adds $15$ dollars per vanilla candle and $20$ dollars per lavender candle: $P = 15x + 20y$.

On the $x$-axis ($y=0$): $3x \le 18$ gives $x \le 6$, and $x \le 10$ from the other limit, so wax binds first: $x=6$, giving $(6,0)$. On the $y$-axis ($x=0$): $2y \le 18$ gives $y \le 9$, and $2y \le 10$ gives $y \le 5$, so curing time binds first: $(0,5)$. The inside vertex is where $3x+2y=18$ meets $x+2y=10$:

$(3x+2y) - (x+2y) = 18 - 10 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4, \quad y = 10 - 4 = 3.$

Evaluate $P = 15x + 20y$ at all four vertices:

$P(0,0)=0, \quad P(6,0)=90, \quad P(0,5)=100, \quad P(4,3)=60+60=120.$

The maximum is $120$ dollars, at $4$ vanilla and $3$ lavender candles.

Substitute the winning plan, $4$ vanilla and $3$ lavender, back into the left side of each original constraint.

$3(4) + 2(3) = 12 + 6 = 18, \qquad 4 + 2(3) = 4 + 6 = 10.$

Both totals hit their limits exactly, so the plan spends all $18$ oz of wax and all $10$ hours of curing time.

That is not a coincidence at an optimum found at an inside vertex: that vertex is where the two slanted constraints cross, so both are tight there by construction. If a resource were instead left over at whatever point maximizes profit, it would mean that resource was never actually forcing a tradeoff, and the true limit on production was coming from elsewhere, a sign that some of that leftover resource could still be put to use, at least up to where the OTHER constraint finally takes over.

Test each of the five points against every one of the four constraints $x \ge 0$, $y \ge 0$, $x \le 3$, $x + y \le 8$, not only the two equations that produced it.

For $(8,0)$: $x = 8 \ge 0$ holds, $y = 0 \ge 0$ holds, $x + y = 8 \le 8$ holds, but

$x = 8 \le 3 \;\text{is false.}$

That single failure is enough to disqualify the point entirely: $(8,0)$ solved $y=0$ and $x+y=8$ correctly, but it was never checked against $x \le 3$, the constraint it actually violates.

Substitute the four valid candidates into $P = 2x + y$ directly.

$P(0,0) = 0, \qquad P(0,8) = 0 + 8 = 8, \qquad P(3,0) = 6 + 0 = 6, \qquad P(3,5) = 2(3) + 5 = 6 + 5 = 11.$

The value at $(3,5)$, $11$, is larger than at every other valid candidate, so it is the true maximum. The list of four is also complete: with $(8,0)$ removed, every remaining pair of boundary lines that actually meets inside the region has been solved.

A boundary line is the equation version of one inequality, holding with equality. Solving two such equations together finds the one point where BOTH of those two conditions hold exactly, and the algebra guarantees nothing more than that: it never looks at any of the region's other constraints.

$\text{solves two boundary equations} \;\ne\; \text{satisfies every constraint}$

A genuine vertex of the feasible region is a point where two boundaries meet AND every other constraint is still satisfied there too. When a problem has more than two boundary lines total, as this one does, some pairs of lines cross outside the region that the OTHER constraints carve out, exactly as $y=0$ and $x+y=8$ cross at $(8,0)$, a point $x \le 3$ rules out. The fix is always the same: after solving a pair of boundary equations, substitute the result into every constraint that was NOT one of the two solved, and only keep the point if all of them hold.

$Q(3,5) = 3 + 5 = 8, \qquad Q(0,8) = 0 + 8 = 8.$

The two values agree, and that is not a coincidence: the segment joining $(3,5)$ and $(0,8)$ is exactly the boundary $x + y = 8$, so every point on it, not only the two endpoints, satisfies $x + y = 8$. Since $Q$ IS $x + y$, the objective equals $8$ at every single point of that edge, because the edge's own equation forces it to.

Evaluating $Q$ at the other two vertices confirms neither beats that: $Q(0,0)=0$ and $Q(3,0)=3$. So the maximum of $Q$ over the whole region is $8$, tied along the entire edge from $(3,5)$ to $(0,8)$, the case where an objective's level lines run parallel to one of the region's own edges.