Writing and interpreting constraints
A budget, a clock and a bag of beads all say “this much and no more”. How do you turn limits like these into mathematics, and then read off which plans are possible and which is best?
Where you are 4 in Systems of linear inequalities
When you reach for it
Kiri makes bracelets and keyrings to sell at a craft stall. She has only so many beads, only so many hours, and she has promised a customer some keyrings. None of these limits tells her what to make. Each one rules some plans out, and the plans that survive all of them are her real choices.
That is the situation this skill is for: two quantities to choose, and several limits on them. A limit written in mathematics is a constraint: an equation or inequality that the chosen quantities must satisfy. A plan that satisfies every constraint is feasible, and on a graph the feasible plans make up the feasible region, the overlap of the regions of all the constraints. A feasible plan is viable when it also makes sense in the situation: no one can make 22.5 bracelets, even when every inequality allows it.
Day 1 is writing the constraints; Day 2 is reading them on a graph. The graphing is graphing a linear inequality and finding where shaded regions overlap, used here rather than retaught.
The procedure
Day 1: turning limits into constraints
Name each unknown as a number. Write “let b be the number of bracelets Kiri makes this week”, not “b is bracelets”: a variable stands for a number.
Write the amount each limit is about. When each item uses a fixed amount, the total is a rate times a count: if a bracelet uses 20 beads and a keyring 10, then b bracelets and k keyrings use 20b + 10k beads. This is the rate-times-quantity reasoning of a linear model, and it is why every constraint here is linear.
Put both sides in the same unit. 20b is (beads per bracelet) \times (bracelets), which is beads, so the other side must be beads too. A limit given in hours is converted to minutes before it is compared with a time counted in minutes.
Choose the sign by asking what the limit allows. A limit is a ceiling, a floor, or a fixed amount.
| The situation says | It means | Sign |
|---|---|---|
| at most, no more than, cannot exceed, up to | a ceiling that may be reached | \le |
| at least, no fewer than, a minimum of | a floor that may be reached | \ge |
| fewer than, less than, under | a ceiling that may not be reached | < |
| more than, over | a floor that may not be reached | > |
| exactly, all of, in full | a fixed amount | = |
The words are a guide; the question underneath is whether anything may be left over. A bag of beads need not be used up, so it gives an inequality. A grant that must be spent in full, or a guest list where everyone gets exactly one meal, gives an equation.
Add the constraints nobody says aloud. Counts and measures cannot be negative, so each quantity gets a constraint such as b \ge 0, unless another constraint already forces it. A count of items (bracelets, laptops, meals) must also be a whole number. A measure (hectares, liters) may be any non-negative number, so 12.5 hectares is a real plan.
Test each constraint on a plan you can judge. “No more than twice as many tablets as laptops” is t \le 2L: 10 laptops should allow up to 20 tablets, and they do. Writing 2t \le L would allow only 5, and the test shows it.
Simplify, but keep the meaning. Dividing 20b + 10k \le 600 by 10, a positive number, gives 2b + k \le 60 with the same sign. The simpler form is easier to graph; the original is the one that says “beads”.
Day 2: reading the constraints
Graph each constraint and find the overlap. Each constraint is a half-plane, solid-edged for \le or \ge and dashed for < or >. The feasible region is the overlap of them all, including the unstated ones.
Find the corners exactly. A corner of the region is a point where two boundary lines meet and every other constraint still holds. Solve the two boundary equations as a system of equations, since corners are often fractions a graph cannot show exactly. A crossing that fails another constraint is not a corner.
Decide which feasible points are viable. For counts, the viable plans are the whole-number points of the region. A fractional corner is not a plan, and the whole points near it must each be tested, since rounding can move a point outside.
Read points back into words. A point on a boundary line uses that limit exactly. A point strictly inside has some of each resource to spare. A corner meets two limits at once.
Compare plans at the corners. When plans are compared by an amount built the same way, such as a profit of 5b + 4k dollars, work it out at each corner. For a straight-line amount like this, over a region closed in on every side, some corner always does at least as well as any other point; this lesson uses that fact without proving it. For counts, if the best corner is not whole, test the whole points near it.
Worked examples
Examples 1 and 2 belong to Day 1, and Examples 3 to 5 to Day 2.
Day 1: Kiri’s craft stall
Kiri sells bracelets and keyrings at a weekend craft stall. Each bracelet uses 20 beads and takes 15 minutes to make; each keyring uses 10 beads and takes 30 minutes. She has 600 beads and at most 12 hours of making time this week, and a customer has already ordered 6 keyrings. Write every constraint on what she can make, and simplify them.
SolutionLet b be the number of bracelets and k the number of keyrings Kiri makes this week.
Beads. The week uses 20b + 10k beads. She cannot use more than she has, and may have some left over, so this is a ceiling:
Time. Making takes 15b + 30k minutes. The limit is in hours, so convert it first: 12 \times 60 = 720 minutes. “At most” includes exactly 720:
The order. She must make at least the 6 promised keyrings, and may make more: k \ge 6. This is a floor, not a fixed amount.
Unstated. b \ge 0; the matching k \ge 0 is already forced by k \ge 6. Both b and k count objects, so both are whole numbers.
Dividing the beads constraint by 10 and the time constraint by 15, both positive, keeps both signs:
Test. 20 bracelets and 10 keyrings use 400 + 100 = 500 beads and 300 + 300 = 600 minutes, so from the story alone Kiri could make them. The constraints agree: 50 \le 60, 40 \le 48 and 10 \ge 6.
Written as 15b + 30k \le 12, the time limit would compare minutes with hours and rule out even a single bracelet. The units are what tell the two apart.
Day 1: A school’s order, one sentence at a time
A school is ordering laptops at 450 dollars each and tablets at 300 dollars each, paid for from a grant of 21{,}000 dollars. Let L be the number of laptops and t the number of tablets. Write a constraint for each condition and say why it has the sign it has.
(a) The order can cost no more than the grant.
(b) The school needs at least 50 devices.
(c) There can be no more than twice as many tablets as laptops.
(d) A different rule: the grant must be spent in full, to the dollar.
(e) A different rule: the order must include more than 15 laptops.
Solution(a) The order costs 450L + 300t dollars, and “no more than” is a ceiling that may be reached: 450L + 300t \le 21{,}000. Every coefficient is a multiple of 150, and dividing by it gives 3L + 2t \le 140.
(b) “At least” is a floor, and exactly 50 devices is allowed: L + t \ge 50.
(c) Twice the number of laptops is 2L, and the tablets may not go above it: t \le 2L, with a variable on each side. The test: 10 laptops allow up to 20 tablets, as the sentence says.
(d) “In full, to the dollar” leaves no room either way, so this is an equation: 3L + 2t = 140. The possible orders are now points on a line, not a region.
(e) “More than 15” does not include 15, so L > 15. Because L counts laptops, this allows the same orders as L \ge 16; for a measured quantity, such as hours, it would not.
The unstated constraints are L \ge 0 and t \ge 0, with both whole numbers.
Day 2: Kiri’s region, and her best plan
Graph the constraints from Example 1 and find the corners of the feasible region. Decide whether 25 bracelets and 10 keyrings, and 20 bracelets and 15 keyrings, are viable plans. Then, with a profit of 5 dollars on each bracelet and 4 dollars on each keyring, find the plan that earns the most. What changes if bracelets earned 8 dollars and keyrings 2?
SolutionThe constraints are 2b + k \le 60, b + 2k \le 48, k \ge 6 and b \ge 0, with b and k whole. Put b across and k up. The beads boundary meets the axes at (30, 0) and (0, 60), the time boundary at (48, 0) and (0, 24), and the origin satisfies both, so both keep the origin side. k \ge 6 keeps the part on or above k = 6. Every sign includes equality, so every boundary is solid.
A: k = 6 meets b = 0, at (0, 6).
B: k = 6 meets 2b + k = 60, so 2b = 54 and the corner is (27, 6). Time: 27 + 2(6) = 39 \le 48.
C: the two resource lines meet where b = 48 - 2k and 2(48 - 2k) + k = 60, so 96 - 3k = 60, k = 12 and b = 24. The corner is (24, 12), and 12 \ge 6.
D: b + 2k = 48 meets b = 0, at (0, 24). Beads: 24 \le 60.
The beads line also crosses the horizontal axis at (30, 0), but k = 0 breaks the promise of 6 keyrings, so that crossing is outside the region.
The two plans. (25, 10) gives 2(25) + 10 = 60 \le 60, 25 + 2(10) = 45 \le 48 and 10 \ge 6, in whole numbers, so it is viable; it lies on the beads boundary, every bead used. (20, 15) gives time 20 + 2(15) = 50, more than 48: not viable, since it needs 300 + 450 = 750 minutes, 30 more than Kiri has.
The best plan. Work out the profit at each corner:
| Corner | Profit 5b + 4k | Profit 8b + 2k |
|---|---|---|
| A(0, 6) | 24 | 12 |
| B(27, 6) | 135 + 24 = 159 | 216 + 12 = 228 |
| C(24, 12) | 120 + 48 = 168 | 192 + 24 = 216 |
| D(0, 24) | 96 | 48 |
With the first prices the best corner is C: 24 bracelets and 12 keyrings, for a profit of 168 dollars. It is whole, so it is viable, and it uses 480 + 120 = 600 beads and 360 + 360 = 720 minutes, both limits met exactly.
With the second prices the best corner is B: 27 bracelets and 6 keyrings, for 228 dollars. Bracelets now pay so much more that Kiri should make only the keyrings she has promised. The region did not change; only the prices did, and a different corner wins.
Day 2: The school’s order, and a corner nobody can buy
The school in Example 2 orders under conditions (a), (b) and (c).
(a) Graph the feasible region and find its corners.
(b) The device line crosses the horizontal axis at (50, 0). Explain why that point is not a corner.
(c) Find the cheapest viable order.
Solution(a) The constraints are 3L + 2t \le 140, L + t \ge 50 and t \le 2L, with L \ge 0, t \ge 0 and both whole. The budget keeps the origin side of its line. The origin fails L + t \ge 50, so the device constraint keeps the far side. The line t = 2L passes through the origin, so test (10, 0) instead: 0 \le 20, true, so the side below the line is kept.
P: budget meets devices. With t = 50 - L, 3L + 2(50 - L) = 140 gives L + 100 = 140, so L = 40 and t = 10. Ratio: 10 \le 80.
Q: budget meets ratio. With t = 2L, 3L + 4L = 140, so L = 20 and t = 40. Devices: 60 \ge 50.
R: devices meet ratio. L + 2L = 50, so \displaystyle L = \frac{50}{3} and \displaystyle t = \frac{100}{3}. Budget: \displaystyle 50 + \frac{200}{3} = \frac{350}{3}, about 116.7, which is at most 140.
(b) (50, 0) is 50 laptops and no tablets. It meets the device minimum and the ratio, but it costs 450 \times 50 = 22{,}500 dollars, more than the grant. Two boundary lines cross there, but the point is outside the region.
(c) The cost is 450L + 300t dollars. At the corners:
| Corner | Cost in dollars |
|---|---|
| P(40, 10) | 18{,}000 + 3{,}000 = 21{,}000 |
| Q(20, 40) | 9{,}000 + 12{,}000 = 21{,}000 |
| \displaystyle R\left(\frac{50}{3}, \frac{100}{3}\right) | 7{,}500 + 10{,}000 = 17{,}500 |
The cheapest corner is R, but nobody can order two thirds of a laptop. Rounding to (17, 33) gives 50 devices, 33 \le 34 and 3(17) + 2(33) = 117 \le 140: viable, at 7{,}650 + 9{,}900 = 17{,}550 dollars. With 16 laptops nothing works: (16, 33) is only 49 devices, and (16, 34) breaks the ratio, since 34 > 32.
Nothing cheaper exists. From t \le 2L and L + t \ge 50, 3L = L + 2L \ge L + t \ge 50, so \displaystyle L \ge \frac{50}{3}, and a whole number of laptops means L \ge 17. The cost can be written 450L + 300t = 150L + 300(L + t), which is at least 150(17) + 300(50) = 17{,}550. The cheapest viable order is 17 laptops and 33 tablets, for 17{,}550 dollars.
Day 2: A constraint that is an equation
Isolde is catering a wedding for 120 guests, each having exactly one meal, vegetarian or meat. At least 30 guests have asked for vegetarian, and the couple want at least 50 meat meals. Ingredients cost 6 dollars for a vegetarian meal and 9 dollars for a meat meal, and she has budgeted at most 960 dollars for ingredients. She charges 30 dollars for a vegetarian meal and 38 dollars for a meat meal. Find every viable plan, and the one with the greatest profit.
SolutionLet v be the number of vegetarian meals and m the number of meat meals. Every guest has exactly one meal, so this limit is an equation: v + m = 120. The others are v \ge 30, m \ge 50 and 6v + 9m \le 960 (that is, 2v + 3m \le 320), with v and m whole numbers.
The equation puts every plan on one line, so the feasible plans form a piece of that line. To find which piece, substitute m = 120 - v into the other constraints.
Ingredients: 6v + 9(120 - v) \le 960, so 1080 - 3v \le 960 and -3v \le -120. Dividing by -3, a negative number, reverses the sign: v \ge 40.
Meat: 120 - v \ge 50, so v \le 70.
Vegetarian: v \ge 30 holds automatically once v \ge 40, so this constraint never bites.
The viable plans are 40 \le v \le 70 with m = 120 - v: that is 70 - 40 + 1 = 31 plans, the whole-number points of the segment from E(40, 80) to F(70, 50).
Each vegetarian meal makes 30 - 6 = 24 dollars and each meat meal 38 - 9 = 29 dollars, so the profit is 24v + 29m. The ends of the segment play the part of corners:
The best plan is 40 vegetarian and 80 meat meals, for a profit of 3280 dollars. On the line the profit is 24v + 29(120 - v) = 3480 - 5v, which falls as v rises, so the end with the fewest vegetarian meals is best; there the ingredients budget is spent exactly, 240 + 720 = 960.
Boundaries
Every item must use a fixed amount. The constraints are linear only because each total is a rate times a count. A bulk discount bends the boundary, and the corner check no longer applies.
The region may be empty. If the school in Example 2 needed at least 80 devices, even the cheapest choice, 80 tablets, would cost 80 \times 300 = 24{,}000 dollars, more than the grant. The model would be saying correctly that the limits cannot all be met.
The region may be unbounded. If every constraint is a floor, the region runs off without end, there may be no largest profit, and checking corners settles nothing.
Strict inequalities leave out their boundary. A corner on a dashed edge is not feasible. For counts, replace L > 15 by L \ge 16; for measures, the best value may be approached but never reached.
Feasible is not yet viable. For counts, the best whole-number plan need not be a corner, and rounding a corner can break a constraint, so each candidate is tested against every one.
Where this goes next
The region drawn here is a system of inequalities read in context, and deciding which of its points make sense is the two-variable form of choosing a reasonable domain for a real situation. When every limit is an equation, the region shrinks to a single point, as in Systems in word problems.
The corner check of Examples 3 to 5 is the first step of optimizing a quantity over a region, where “which plan is best” becomes the whole subject.
Exercises
Exercises 1 to 7 go with Day 1 (Examples 1 and 2), Exercises 8 to 16 with Day 2 (Examples 3 to 5), and the reasoning group draws on both.
Day 1: one sentence, one constraint
Write a constraint for each sentence and say why it has the sign it has.
- Hyacinth makes bracelets and keyrings. Each bracelet uses 15 beads and each keyring 10, and she has a bag of 450 beads. Use b bracelets and k keyrings.
- A farmer has contracted to deliver at least 300 tonnes of potatoes, and each hectare of potatoes yields 30 tonnes. Use p hectares.
- A school buying laptops at 500 dollars and tablets at 250 dollars must use the whole grant of 30{,}000 dollars, no more and no less.
Day 1: a whole situation
- Galen farms 50 hectares and will plant corn and beans, c hectares of corn and s hectares of beans; he need not plant it all. Corn needs 4 megaliters of water per hectare and beans 2.5 megaliters per hectare, and he may use at most 160 megaliters. He wants at least 8 hectares of beans. Write every constraint, including any unstated ones, and say whether c and s must be whole numbers.
- Hyacinth, from Exercise 1, takes 10 minutes to make a bracelet and 15 minutes to make a keyring, and has 8 hours for making. (a) Write the time constraint. (b) Using it and the beads constraint, decide which plans are possible: 20 bracelets and 10 keyrings; 10 bracelets and 25 keyrings; 26 bracelets and 8 keyrings; 6 bracelets and 30 keyrings.
- A school’s order of L laptops and t tablets is governed by 600L + 400t \le 24{,}000, L + t \ge 45, L \ge 10 and t \le 3L. (a) Say in words what each constraint means. (b) Is an order of 20 laptops and 30 tablets allowed?
- A school decides that at least a third of the devices it orders must be laptops. Write this as a constraint on L and t, and simplify it.
Day 2: graphs and corners
- Graph Hyacinth’s constraints from Exercises 1 and 5, find the corners of the feasible region, and decide whether the corner where the two resource lines meet is a viable plan. Are (16, 21) and (15, 22) viable?
- Find the corners of Galen’s feasible region from Exercise 4. Which of them are viable plans?
- Find the corners of the feasible region of the school in Exercise 6. The line L + t = 45 crosses the horizontal axis at (45, 0); explain why that point is not a corner.
Day 2: viable or not
- For Kiri’s stall in Examples 1 and 3, decide whether each is a viable plan, and if not, which condition fails: (26, 8), (10, 20), (22.5, 10), (29, 2).
- A caterer, Leontine, must serve exactly 90 meals at an event. Ingredients cost 5 dollars for a vegetarian meal and 8 for a meat meal, she has at most 630 dollars for ingredients, and at least 25 meals must be meat. Find the range of possible numbers of vegetarian meals, and how many viable plans there are.
Day 2: comparing options
- Hyacinth (Exercise 8) makes a profit of 4 dollars on a bracelet and 3 dollars on a keyring. Find the profit at each corner, and the whole-number plan that earns the most.
- Galen (Exercise 9) expects a profit of 900 dollars per hectare of corn and 600 dollars per hectare of beans. Which plan earns the most, and how much?
- Which viable order gives the school in Exercise 10 the most devices? Explain why rounding the best corner does not work.
- For the event in Exercise 12, Leontine makes a profit of 14 dollars on a vegetarian meal and 12 dollars on a meat meal. Which plan earns the most, and how much?
Reasoning
- No problem ever says “Kiri cannot make a negative number of bracelets”. Explain what goes wrong in Example 3 if b \ge 0 is left out.
- The best plan for Galen’s farm in Exercise 14 has fractional coordinates, and that is fine. The best corner for Hyacinth in Exercise 13 has fractional coordinates, and that is not. Explain the difference.
- Explain why a constraint that is an equation turns the feasible region into a segment, as in Example 5, and why the ends of the segment are the places to check.
- Kiri’s profit is 5b + 4k, with both rates positive. Explain why the most profitable point of her region can never lie strictly inside it.
Answers
- 15b + 10k \le 450, or 3b + 2k \le 90. A ceiling: beads may be left over.
- 30p \ge 300, that is p \ge 10. A floor, in tonnes on both sides.
- 500L + 250t = 30{,}000, or 2L + t = 120. An equation: “no more and no less”.
- c + s \le 50; 4c + 2.5s \le 160, or 8c + 5s \le 320; s \ge 8; c \ge 0, with s \ge 0 forced by s \ge 8. Not whole numbers: they measure land, and part of a hectare can be planted.
- (a) 10b + 15k \le 480, since 8 hours is 480 minutes; that is 2b + 3k \le 96. (b) With beads 3b + 2k \le 90: (20, 10) gives 80 and 70, possible; (10, 25) gives 80 and 95, possible; (26, 8) uses 94 beads, not possible; (6, 30) needs time 102, not possible.
- (a) The order costs at most 24{,}000 dollars; it has at least 45 devices; at least 10 are laptops; there are at most three tablets per laptop. (b) Yes: 12{,}000 + 12{,}000 = 24{,}000, 50 \ge 45, 20 \ge 10 and 30 \le 60.
- \displaystyle L \ge \frac{1}{3}(L + t). Multiplying by 3, 3L \ge L + t, so t \le 2L.
- Corners (0, 0), (30, 0), \displaystyle \left(\frac{78}{5}, \frac{108}{5}\right) = (15.6, 21.6) and (0, 32). The corner (15.6, 21.6) is not viable: items are whole. (16, 21) is viable (beads 90, time 95), and so is (15, 22) (beads 89, time 96).
- Corners (0, 8), (35, 8), \displaystyle \left(\frac{70}{3}, \frac{80}{3}\right) and (0, 50). All four are viable: hectares need not be whole.
- Corners \displaystyle \left(\frac{45}{4}, \frac{135}{4}\right) = (11.25, 33.75), \displaystyle \left(\frac{40}{3}, 40\right) and (30, 15); L \ge 10 never touches the region. The point (45, 0) costs 600 \times 45 = 27{,}000 dollars, more than the grant.
- (26, 8) is viable: 60 \le 60, 42 \le 48, 8 \ge 6. (10, 20) is not: time gives 50 > 48. (22.5, 10) meets every inequality but is not viable: half a bracelet. (29, 2) is not: fewer than the 6 promised keyrings.
- 30 \le v \le 65, with m = 90 - v: 36 viable plans. From 720 - 3v \le 630, v \ge 30; from 90 - v \ge 25, v \le 65.
- (0, 0) gives 0, (30, 0) gives 120, (15.6, 21.6) gives 127.2, (0, 32) gives 96. The best corner is not whole; (16, 21) earns 64 + 63 = 127 dollars, and no whole plan can earn more, since whole plans earn whole dollars and none can pass 127.2.
- \displaystyle \left(\frac{70}{3}, \frac{80}{3}\right), about 23.3 hectares of corn and 26.7 of beans, for 37{,}000 dollars. The other corners give 4800, 36{,}300 and 30{,}000.
- 14 laptops and 39 tablets, 53 devices. The best corner \displaystyle \left(\frac{40}{3}, 40\right) has 53.3 devices, so no order has more than 53; rounding it to (13, 40) breaks t \le 3L, since 40 > 39. (14, 39) costs exactly 24{,}000 dollars and has 39 \le 42.
- 65 vegetarian and 25 meat meals, for 1210 dollars. The profit is 14v + 12(90 - v) = 1080 + 2v, which rises with v; the other end, (30, 60), gives 1140.
- Without b \ge 0 the region runs off to the left without end: (-10, 6) satisfies the other three constraints. The algebra would accept “minus ten bracelets”, and corners A and D would vanish.
- Galen’s quantities measure land, and 23.3 hectares is a real field, so every feasible point is viable. Hyacinth’s count objects, and 15.6 bracelets cannot be made, so her best corner must be replaced by a nearby whole point that is tested.
- An equation allows only the points on its line. Each other constraint keeps the part of the line on one side of a crossing, so what remains lies between two crossings: a segment. Along it the compared amount is a linear function of one variable, so it rises or falls steadily, and its largest and smallest values are at the ends.
- From a point strictly inside, Kiri can move a little to the right and stay in the region. That raises b, and so raises 5b + 4k, since 5 is positive. So an inside point is never the best.