Testing every corner, and the extra check an unbounded region forces

What to assume they know

• The feasible region and how to draw it, from the first topic of this module
• Theorem 1, Theorem 2 and the Remark on the unbounded case, from the second
• Finding where two lines cross by solving their equations together
• Substituting a pair of numbers into a linear expression, including with negative coefficients
• Graphing a strict linear inequality, and knowing that its boundary is excluded

What they should be able to do

• Carry out the chapter's four printed steps in order on a bounded problem, and say what each step is for
• Find corner coordinates either by reading them off the drawing or by solving two equations, and say when each way is safe
• Evaluate the objective at every corner and name the largest and the smallest of those values
• Say why, on a bounded region, those two values settle the problem outright
• Say why, on an unbounded region, they settle nothing until one more test is run
• State the extra test in both directions, for a largest and for a smallest value, and carry it out on a drawing
• Explain why the test line is drawn dashed and every constraint line is drawn solid
• Conclude correctly when the test finds common points, and when it finds none
• Name the five shapes an answer to a problem in this chapter can take
• Sort a set of stated problems by which of those five shapes they land in

Where it usually goes wrong

• "Once I have the largest corner value I am finished." Only if the region is bounded. On an unbounded region that number is a candidate and nothing more, and Example 4 spends a page and a half showing a candidate failing.
• "Unbounded means there is no answer." Three of the ten exercise items have unbounded regions and two of them have perfectly good answers. Unbounded means test, not give up.
• "The test inequality uses the same sign as the objective's constraint." It uses a strict inequality, in the improving direction: strictly greater when you are checking a largest value, strictly smaller when you are checking a smallest one. Getting the direction backwards reverses the conclusion.
• "If the test line touches the region, the test has failed." The test line always touches the region — it passes through the very corner you are testing. The question is whether the open half plane strictly beyond it contains any point of the region. Fig 12.5 makes this trap visible and the chapter does not warn about it.
• "The test line is a constraint I forgot." It is not part of the problem at all. It is drawn to answer a question about the answer, and it is dashed for exactly that reason.
• "Corner coordinates can always be read off the graph." Exercise 12.1 item three has a corner at nineteenths. Solve the two equations whenever the crossing does not land on the ruling.
• "Common points found means the candidate is wrong by a little." It means there is no extreme value at all in that direction. The conclusion is not a corrected number; it is that no number exists.
• "A negative coefficient in the objective means something has gone wrong." Example 4 and exercise item two both have one, and both are ordinary problems. What a negative coefficient changes is which corner wins, not whether the method applies.
• "The chapter's unanswered question is optional." It is the second half of the test — the direction Example 4 never demonstrates. A student who skips it has seen the test fail and never seen it pass.

Questions to check understanding

• Solve a stated problem on a bounded region by the four printed steps, showing the corner table
• Find a corner that does not land on the ruling, by solving two equations
• Say, for a stated problem, whether step three finishes it or whether the extra test is needed, and justify the answer from the region alone
• Write down the test inequality for a stated candidate value, in the correct direction
• Say whether a drawn test line should be solid or dashed, and why
• Given a drawing in which the test half plane meets the region, state the conclusion in words
• Show that a stated problem has no maximum — the form of Exercise 12.1 item nine
• Show that the smallest value of a stated problem is attained at more than two separate points — the form of Exercise 12.1 item six
• Classify a stated problem into one of the five outcome shapes before solving it

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The printed steps (§12.2.2, Part II p. 399). Read off a three-hundred-dot-per-inch the printed page. Step one: find the region and its corners, either by reading them straight off the drawing or by solving together the equations of whichever pair of lines meets at that corner. Step two: evaluate the objective at each corner, and give names to the largest and the smallest of the values found. Step three, part one: when the region is bounded, those two are the answer. Step three, part two: introduces the unbounded case and stops at a colon. Step four, parts one and two: carries the whole content of the unbounded case. The numbering is broken there — see Notes.
• Example 1 and Fig 12.2 (Part II pp. 399–400). Maximise a two-term objective with coefficients four and one, under an upper limit of fifty on the plain sum and of ninety on a weighted sum, both variables non-negative. The corners are the origin; thirty and zero; twenty and thirty; and zero and fifty. Verified: the objective takes zero, one hundred and twenty, one hundred and ten, and fifty at those four points, so the largest is one hundred and twenty at thirty and zero. The region is bounded, so step three part one ends it. This is the cleanest drill in the chapter and it should carry section 5. Note that the winner sits on the horizontal axis, not at the interior corner a student expects to win.
• Example 2 and Fig 12.3 (Part II pp. 400–401). Minimise an objective with coefficients two hundred and five hundred, under a lower limit of ten on one weighted sum and an upper limit of twenty-four on another, both variables non-negative. Verified: the three corners are zero and five, four and three, and zero and six; the objective takes two thousand five hundred, two thousand three hundred, and three thousand there, so the smallest is two thousand three hundred at four and three. Bounded again. A drawing note: read off a three-hundred-dot-per-inch the printed page, the region is a very thin triangle wedged against the vertical axis between the heights five and six, and the printed vertical scale omits both those numerals, printing the corner letters in their place. Both boundary lines are solid. A redraw should restore the numerals and open the wedge up, because at printed scale a student cannot see that there are three distinct corners.
• What step one's two routes cost (not in the book). The chapter offers reading the corner off the drawing or solving the two equations. Verified against Exercise 12.1 item three: the corner there is at twenty over nineteen and forty-five over nineteen, which no drawing will ever yield. Reading off the picture is safe only when the crossing lands on the ruling; otherwise solve. The chapter presents the two routes as equals and never warns which is which.
• Example 4 and Fig 12.5, the unbounded case (Part II pp. 402–403). Minimise an objective whose first coefficient is negative fifty and whose second is twenty, under three constraints plus non-negativity. Verified: the corners are zero and five, zero and three, one and zero, and six and zero; the objective takes one hundred, sixty, negative fifty and negative three hundred there, so the smallest found is negative three hundred at six and zero. The region is unbounded, so that number is only a candidate. The chapter then graphs the strict inequality that asks for values below negative three hundred, simplifies it by dividing through by ten, and reads off the figure that it does meet the region — so there is no minimum. Verified independently by arithmetic added here: the point with first coordinate ten and second coordinate eight thirds satisfies all three constraints, both non-negative conditions, and the strict inequality, since the objective there is about negative four hundred and forty-seven. So common points genuinely exist and the chapter's conclusion is right.
• The dashed line (Fig 12.5, Part II p. 402). Read off a three-hundred-dot-per-inch the printed page. The three constraint boundaries are solid; the test line is dashed, and it is the only dashed line anywhere in the chapter. The figure also carries two shading tones, the pale one the whole feasible region and the darker one the overlap being tested — which is the visual proof that the two meet. Verified reading: the test line is dashed because the inequality it comes from is strict and its own points are excluded, where every constraint permits equality and keeps its boundary. The chapter never says this — the words solid and dashed appear nowhere in it — and it is the single most important unwritten fact in this topic.
• The dashed line passes through the winning corner (Fig 12.5, Part II p. 402). Verified: the test line is the set where the objective equals negative three hundred exactly, so it passes through the corner at six and zero. That is worth pointing out: the corner itself is on the line and therefore not in the open half plane, and the test is about whether anything beyond it survives. Students who see the line touching the corner conclude the test has already succeeded.
• The question the chapter asks and does not answer (Part II p. 403). Immediately after finishing Example 4 the chapter asks whether the same objective has a largest value of one hundred at the corner zero and five, tells the reader to test the opposite open half plane, and closes with a bare why. No answer is printed anywhere. Verified by working added here: the test asks for the objective to exceed one hundred, which after dividing through by ten means the second variable must exceed five plus two and a half times the first; but one of the constraints caps the second variable at five plus twice the first. The two can hold at once only when the first variable is negative, and it is not allowed to be. So the open half plane meets the region nowhere, and one hundred is the maximum, attained at zero and five. It is the chapter's only unanswered question and it is exactly the second direction of the test.
• Example 5 and Fig 12.6 (Part II p. 403). The fifth outcome: no region at all, so the method never starts. Verified: on the triangle the upper limit allows, the plain sum of the two variables cannot exceed five, and the other constraint demands at least eight. Handled in full by the first topic of this module; named here only so the taxonomy in section 11 is complete.
• The five shapes an answer can take (not in the book). One: bounded region, one winning corner — Examples 1 and 2. Two: bounded region, two winning corners and the whole segment between them — Example 3. Three: unbounded region, the extreme value found is not attained — the minimum in Example 4. Four: unbounded region, the extreme value found is attained — the maximum in Example 4, the one the chapter leaves as a question. Five: no feasible region at all — Example 5. The chapter never assembles this list; the Remarks on Part II p. 403 gather only the first two.
• Exercise 12.1, all ten items, sorted by outcome (Part II pp. 403–404). Verified by working added here from the printed constraints; the chapter prints no answers. Item one, bounded, largest sixteen at zero and four. Item two, bounded, smallest negative twelve at four and zero — the only item with a negative coefficient in the objective. Item three, bounded, largest two hundred and thirty-five nineteenths at twenty nineteenths and forty-five nineteenths — the only item whose answer is not a whole number, and the one that proves step one's second route is not optional. Item four, unbounded, smallest seven at three halves and one half, and the value is attained. Item five, bounded, largest eighteen at four and three. Item six, unbounded, smallest six along the whole segment from zero and three to six and zero, which is what its own instruction line asks the student to show. Item seven, bounded, smallest three hundred at one corner and largest six hundred at two. Item eight, bounded, smallest one hundred at two corners and largest four hundred at one. Item nine, unbounded, and there is no maximum — the largest corner value is one, and the open half plane above it meets the region, for instance at the point three and ten. Item ten, no feasible region. So the exercise lands six items in shape one, two in shape two, one in shape three and one in shape five, with item four in shape four.

Figures to have open

• A step diagram for section 1 showing the printed numbering and the corrected numbering side by side. The content is the chapter's Part II p. 399; the correction is added here and must be labelled as such.
• A two-column corner-finding panel for section 2, the second column carrying the simultaneous equations of Exercise 12.1 item three worked to nineteenths. Entirely added here; the chapter works no corner algebraically anywhere.
• A corner-and-value table for sections 3 and 5, built with the repo's DataTable component, from Fig 12.2's numbers on Part II p. 400.
• A redraw of Fig 12.2 (Part II p. 400) for section 5. Both boundaries solid; four corners; restore the two horizontal axis numerals the printed figure drops in favour of point labels.
• A redraw of Fig 12.3 (Part II p. 400) for section 6, rescaled. The printed figure's region is too thin to read and its vertical scale omits the two numerals the corners sit at. Read the corner coordinates off the printed page listed in Notes.
• A solid-against-dashed pair for section 8, from Fig 12.5 (Part II p. 402). The distinction is the chapter's drawing; the labelling is added here, since the chapter names neither line style.
• A redraw of Fig 12.5 (Part II p. 402) for section 9, keeping both shading tones and the dashed line, and making visible that the dashed line passes through the tested corner. This is the chapter's most information-dense figure.
• A five-card outcome panel for section 11 and a ten-row sorted table for section 12. Both entirely added here; the chapter neither assembles the taxonomy nor sorts its own exercise.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 12 "Linear Programming", §12.2.2, the Corner Point Method and its four printed steps, Part II p. 399
• Example 1 with Fig 12.2 and its corner table, Part II pp. 399–400
• Example 2 with Fig 12.3 and its corner table, Part II pp. 400–401
• Example 4 with Fig 12.5, its corner table, and the open-half-plane test, Part II pp. 402–403
• The unanswered question about the maximum, Part II p. 403
• Example 5 with Fig 12.6, and Remarks items (i) and (ii), Part II p. 403
• Exercise 12.1, all ten items, Part II pp. 403–404