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Room for everyone
Bound, then build the best
A good design shows what can work. A proof of the best must also show why nothing better can work.
A shared garden begins with people, space and a clear purpose for the plan.
Mina and Rowan are helping plan a community garden: places to grow flowers, meet friends and move supplies. Sprout is their coach. They begin with a small rectangular plot.
Mina “This outline fits our edging. Does that make it the best one?”
CKSprout “What does your outline establish, and what still needs a reason?”
Rowan “It shows one design is possible. We haven’t ruled out a bigger area.”
CKSTEM Thinking CompassStuck? Find your next move.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
One working plan is a beginning
Use all 14 metres of edging around the four sides of one rectangle. Each side length must be a positive whole number of metres. There are no gaps or overlaps in the edging; joining it loses no length.
What is the largest possible area, and which side lengths achieve it? Explain why a larger area cannot work.
The sketch shows four sides, not chosen dimensions. All fourteen metres must be used; sides are positive whole metres.Try with the CKSTEM Thinking Compass
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What is the largest possible area, and which side lengths achieve it? Explain why a larger area cannot work.
One working plan is a beginning
Use all 14 metres of edging around the four sides of one rectangle. Each side length must be a positive whole number of metres. There are no gaps or overlaps in the edging; joining it loses no length.
The sketch shows four sides, not chosen dimensions. All fourteen metres must be used; sides are positive whole metres.
Your answer
Enter the largest area and its shorter and longer sides. Explain why no other allowed plot has a greater area.
These quantities fit. Your bound and the construction reaching it still need to be justified against every original condition.
Break it into steps
Step 1 of 4
What do two neighbouring side lengths add to?
Half the boundary is 7 metres.
Step 2 of 4
What is the largest allowed area?
The complete comparison has maximum 12, attained by 3 by 4.
Step 3 of 4
What is the shorter side of a maximum-area whole-metre plot?
The greatest area uses a shorter side of 3 m.
Step 4 of 4
What is its longer side?
The other side is 4 m, and 3×4=12.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
What would you need to show about other allowed designs before calling one the largest?
Show the answer and reasoning: the small plot
Answer: The largest area is 12 m², reached by a 3 m by 4 m rectangle.
Rowan “The four sides use 14 metres, so two neighbouring sides add to seven. I can cover every whole-number choice.”
CKSprout “How will that comparison support your claim?”
Mina “No row will have area above twelve. Then we can check that our three-by-four plot really reaches twelve.”
Every positive whole-number pair adding to seven
One side
Other side
Area
1 m
6 m
6 m²
2 m
5 m
10 m²
3 m
4 m
12 m²
4 m
3 m
12 m²
5 m
2 m
10 m²
6 m
1 m
6 m²
The reversed pairs repeat areas, but no allowed pair is omitted. The greatest entry is twelve.
Exclude a better result
Every allowed pair has area at most 12 m².
Reach the bound legally
3 m × 4 m gives 12 m², using all 14 m.
Each unit square is one square metre. The picture is an allowed example; the complete comparison supplies the upper bound.
Check: 3 + 4 + 3 + 4 = 14 metres, and 3 × 4 = 12 square metres. The complete table proves that every allowed area is at most 12. The actual rectangle proves that 12 is possible.
Bound, then build the best
To prove a largest value, show that every allowed result is at most your bound. Then construct an allowed result reaching it. For a smallest value, show that every result is at least your bound, then reach that lower limit.
Keep the two jobs separate: exclude a better result and build a matching one. Check the construction against every original rule. A bound alone may not be reachable; one excellent example may still be beaten.
Before choosing a method, be precise about what “best” means here: biggest area, fewest moves, shortest time or another stated goal.
What if the lengths need not be whole numbers?
A new rectangular plot must use all 24 metres of edging on its four sides. In this ideal plan, side lengths may be any positive numbers, including fractions or decimals. Joining the edging loses no length. A square is allowed.
Find the largest possible area and an allowed design reaching it. Your reason must cover fractional lengths too, not only a whole-number table.
The new total is twenty-four metres. Fractional positive lengths and a square are allowed; the sketch does not choose them.Try with the CKSTEM Thinking Compass
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Find the largest possible area and an allowed design reaching it. Your reason must cover fractional lengths too, not only a whole-number table.
What if the lengths need not be whole numbers?
A new rectangular plot must use all 24 metres of edging on its four sides. In this ideal plan, side lengths may be any positive numbers, including fractions or decimals. Joining the edging loses no length. A square is allowed.
The new total is twenty-four metres. Fractional positive lengths and a square are allowed; the sketch does not choose them.
Your answer
Enter the largest area and both neighbouring sides. Your bound must cover all allowed positive lengths, including fractions.
These quantities fit. Your bound and the construction reaching it still need to be justified against every original condition.
Break it into steps
Step 1 of 4
What side length would the allowed square have?
The square has side 6 metres.
Step 2 of 4
What is the largest area when any positive lengths are allowed?
The product is 36 minus the square of each side’s distance from 6, so it cannot exceed 36.
Step 3 of 4
What is one side of the maximum-area plot?
Each side of the attaining square is 6 m.
Step 4 of 4
What is its neighbouring side?
Each side of the attaining square is 6 m.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: all positive lengths
Answer: The maximum is 36 m², achieved by the 6 m by 6 m square.
Balance the factors
Two neighbouring sides add to 12 metres. Compare a 4 m by 8 m rectangle with a square whose sides are both 6 metres.
Mina “I can cut a two-by-four strip from the end and turn it under the remaining piece.”
CKSprout “What stays unchanged when you rearrange those pieces?”
Rowan “Their total area. They fit inside the six-by-six square, but leave a two-by-two corner missing.”
Cut the rectangle
Turn the strip
The two coloured pieces keep their area. Together they fill the six-by-six square except for a two-by-two corner; this example motivates the general identity below.
The example suggests the reason. Now cover every pair of positive side lengths. Let \(d\) be how far the shorter side falls below 6. The longer side is the same distance above 6, because their sum stays 12.
Here \(0\leq d\lt6\). A square has \(d=0\), so it loses no corner area. Every unequal pair loses a positive square’s worth.
Exclude a better result
Every allowed area is 36 − d² ≤ 36.
Reach the bound legally
6 m × 6 m reaches 36 m², with d = 0.
The equality case must be allowed. Here the six-by-six square uses exactly twenty-four metres of edging.
Check: four sides of 6 metres use exactly 24 metres, and the square is allowed. The formula supplies the universal bound; this allowed square reaches it.
The same reasoning works for any fixed total of two positive factors. Let \(m\) be half that total and \(d\) their distance from it. The factors are \(m-d\) and \(m+d\), and
\[(m-d)(m+d)=m^2-d^2.\]
The smallest allowed distance from the middle gives the largest product. First check that the sum really is fixed and that the proposed factors are allowed.
Can every bound be reached?
A visitor asks for an oblong plot: its side lengths must differ. That small change matters.
Use all 24 metres on the four sides of a rectangle, with no joining loss. Both side lengths must be positive and unequal.
Case A: in the ideal plan, lengths may be any positive numbers, including fractions and decimals. Case B: both lengths must be positive whole numbers of metres.
For each case, is there a largest possible area? If so, find it and give an allowed design. If not, explain why no allowed design is best.
Case A
Any positive lengths
Unequal sides; fractional lengths allowed
Case B
Positive whole metres
Unequal sides
Both cases use the same twenty-four metres. The allowed side lengths differ, so their best-value conclusions may differ.Try with the CKSTEM Thinking Compass
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For each case, is there a largest possible area? If so, find it and give an allowed design. If not, explain why no allowed design is best.
Can every bound be reached?
A visitor asks for an oblong plot: its side lengths must differ. That small change matters.
Use all 24 metres on the four sides of a rectangle, with no joining loss. Both side lengths must be positive and unequal.
Case A: in the ideal plan, lengths may be any positive numbers, including fractions and decimals. Case B: both lengths must be positive whole numbers of metres.
Case A
Any positive lengths
Unequal sides; fractional lengths allowed
Case B
Positive whole metres
Unequal sides
Both cases use the same twenty-four metres. The allowed side lengths differ, so their best-value conclusions may differ.
Your answer
Decide whether Case A has a largest area, then enter Case B’s greatest area and both attaining sides. Explain both conclusions separately.
Your two conclusions and the whole-number dimensions fit. Explain both cases and check them against every original condition.
Break it into steps
Step 1 of 5
In Case B, what is the smallest positive whole difference between 6 and an allowed shorter side?
The nearest allowed shorter side is 5, a gap of 1 metre.
Step 2 of 5
Does Case A have a largest possible area?
No. Bringing the sides closer to six while keeping them unequal always produces a larger allowed area.
Step 3 of 5
What is the largest area in Case B?
Case B has maximum 35 at 5 by 7.
Step 4 of 5
In Case B, what is the shorter side of a best plot?
The shorter side is 5 m.
Step 5 of 5
In Case B, what is its longer side?
The other side is 7 m, giving 35 square metres.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the two domains
Answer:Case A has no largest area. Areas can approach 36 m², but no allowed plot reaches it. Case B has maximum 35 m², achieved by 5 m by 7 m.
Rowan “Five by seven gives thirty-five. But five-point-nine by six-point-one gives more, without being a square.”
CKSprout “What would show whether every allowed proposal can be improved?”
Mina “Keep it unequal but halve its distance from six. That makes the missing corner smaller.”
4 m × 8 m
Area 32 m²
5 m × 7 m
Area 35 m²
5.9 m × 6.1 m
Area 35.99 m²
5.99 m × 6.01 m
Area 35.9999 m²
Closer unequal sides give areas closer to thirty-six. These examples suggest improvement; the argument for every positive gap is still needed.
Those examples do not prove the general claim. In Case A, any allowed plot has a positive gap \(d\) and area \(36-d^2\). Replace its side lengths by \(6-d/2\) and \(6+d/2\). They stay positive, unequal and add to 12.
Every allowed plot has a better allowed plot. So there is no largest one. The ceiling 36 is useful, but the only way to reach it would be the forbidden square.
Case B changes the answer. Its shorter side is a whole number below 6, so it is at most 5. Its gap \(d\) is therefore at least 1. The area is at most \(36-1^2=35\). A 5 by 7 rectangle reaches 35.
Check: 5 + 7 + 5 + 7 = 24, and the whole-number sides differ. Halving its gap would produce 5.5 and 6.5, which are allowed in A but not B. The same upper ceiling does not make the two allowed domains identical.
A wall changes the resource rule
Another plot will sit against all or part of a 13 m straight wall. The side along the wall needs no edging and cannot exceed 13 m. Use all 30 m of edging on the other three sides, with no joining loss.
The two sides extending out from the wall have equal length \(w\); the side along the wall has length \(l\). Both are positive whole numbers of metres. Part of the wall may remain unused.
What is the largest area allowed by all these conditions, and which dimensions achieve it? Prove that no other allowed design is better.
The grey wall replaces one side of edging. The two sides extending out from it both need edging; unused wall is allowed.Try with the CKSTEM Thinking Compass
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What is the largest area allowed by all these conditions, and which dimensions achieve it? Prove that no other allowed design is better.
A wall changes the resource rule
Another plot will sit against all or part of a 13 m straight wall. The side along the wall needs no edging and cannot exceed 13 m. Use all 30 m of edging on the other three sides, with no joining loss.
The two sides extending out from the wall have equal length \(w\); the side along the wall has length \(l\). Both are positive whole numbers of metres. Part of the wall may remain unused.
The grey wall replaces one side of edging. The two sides extending out from it both need edging; unused wall is allowed.
Your answer
Enter the greatest area and its oriented dimensions. Check the wall limit, all edging and why every other whole width is worse.
These quantities fit. Your bound and the construction reaching it still need to be justified against every original condition.
Break it into steps
Step 1 of 5
What is the least possible whole width out from the wall?
The length at most 13 forces two widths at least 17, so the whole width is at least 9.
Step 2 of 5
What is the greatest possible whole width?
Two widths must use less than 30 metres, so a whole width is at most 14.
Step 3 of 5
What is the largest allowed wall-plot area?
The complete list gives 108 at 9 out and 12 along.
Step 4 of 5
What is the best width out from the wall?
The best whole width extends 9 m out from the wall.
Step 5 of 5
What is its length along the wall?
The side along the wall is 12 m; it fits within the 13 m wall.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the wall plot
Answer: The maximum is 108 m²: extend 9 m out from the wall and use 12 m along it.
Two widths and one length use the edging:
\[2w+l=30.\]
Rowan “I can balance the two quantities that add to thirty: twice the width, and the length. Their product is twice the area.”
CKSprout “What would make that bound a finished answer?”
Mina “An allowed rectangle reaching it. Equal quantities would give a width of seven-and-a-half and a length of fifteen. Neither fits our rules.”
This is a valid upper bound, but equality would require a fractional width and more wall than we have. It leaves the actual best whole-metre design undecided. We need a sharper comparison.
Mina “The ten-by-ten square works, but its area is only a hundred. Width plus length is not fixed this time.”
CKSprout “How will you cover the allowed choices without relying on that square guess?”
Rowan “The wall and the edging limit the width. I’ll find that range, then check every whole width in it.”
Because \(l\leq13\), the two widths use at least 17 metres. A whole width must be at least 9 metres. Because \(l\) is positive, \(2w\lt30\), so a whole width is at most 14 metres.
Make an organized list
Fix the width in order from 9 through 14. Each width forces exactly one length, \(l=30-2w\), so the list covers every allowed rectangle. A direct algebraic comparison is possible too; six rows make the wall and whole-number checks easy to see.
Every allowed whole width, 9 through 14
Out from wall: w
Along wall: l
Area
9 m
12 m
108 m²
10 m
10 m
100 m²
11 m
8 m
88 m²
12 m
6 m
72 m²
13 m
4 m
52 m²
14 m
2 m
28 m²
Each row uses two widths plus one length totalling thirty. The wall limits the length; the complete table proves the greatest allowed area is 108.
Bound, then build the best asks for both pieces of evidence. Every allowed row has area at most 108. Now build the row that reaches 108.
Exclude a better result
Every permitted whole-metre design has area at most 108 m².
Reach the bound legally
9 m out × 12 m along uses 30 m edging and fits the 13 m wall.
One square is one square metre. All 108 squares form an allowed rectangle; one metre of existing wall stays unused.
Check: the edging is 9 + 12 + 9 = 30 metres. The 12 m side fits along the 13 m wall, leaving 1 m unused. Both dimensions are positive whole numbers, and 9 × 12 = 108. Turning the labels around would change which side needs edging twice, so that is a different resource calculation.
Revisit Make an organized list when a bounded set of choices must be covered completely.
Can the shortest-looking route actually work?
The garden team uses two cart tokens to check a narrow passing place before deciding how the route should work.
The map shows all allowed places and connections. Places 0, 1, 2, 3 and 4 form a straight path; bay A connects only to 2. The connections work in either direction.
Red token R starts at 0, and blue token B starts at 4. They must exchange ends: R finishes at 4 and B at 0. Every other place starts empty.
One move sends one token along one connection into an empty place. Tokens cannot jump, share a place, swap simultaneously or leave the shown map.
What is the fewest possible moves? Give a legal sequence reaching it and explain why fewer moves cannot work.
Start
Required finish
The full map stays visible in both states. R and B identify the tokens as well as colour; dots are empty places, not extra tokens.Try with the CKSTEM Thinking Compass
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What is the fewest possible moves? Give a legal sequence reaching it and explain why fewer moves cannot work.
Can the shortest-looking route actually work?
The garden team uses two cart tokens to check a narrow passing place before deciding how the route should work.
The map shows all allowed places and connections. Places 0, 1, 2, 3 and 4 form a straight path; bay A connects only to 2. The connections work in either direction.
Red token R starts at 0, and blue token B starts at 4. They must exchange ends: R finishes at 4 and B at 0. Every other place starts empty.
One move sends one token along one connection into an empty place. Tokens cannot jump, share a place, swap simultaneously or leave the shown map.
Start
Required finish
The full map stays visible in both states. R and B identify the tokens as well as colour; dots are empty places, not extra tokens.
Your answer
Enter the minimum. Construct your own full route and justify the lower bound; the checker does not assess the route or proof.
The number fits. Your reasoning, legal construction and check of every rule remain unscored.
Break it into steps
Step 1 of 3
How many main-path moves do the two net journeys need altogether?
Four plus four gives a lower bound of eight main-path moves.
Step 2 of 3
How many extra moves does entering and leaving the bay require?
Entry and exit add at least two moves. This is still not a constructed route.
Step 3 of 3
What is the fewest possible total moves?
A legal ten-move route matches the lower bound.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the passing place
Answer: The minimum is 10 moves. The construction below reaches it, and the lower-bound argument rules out every shorter route.
Rowan “Each token must cross four main-path connections. That means at least eight moves altogether.”
Mina “But adding their distances doesn’t tell us how they get past each other.”
CKSprout “What would any successful route have to change?”
Find an invariant
Rowan “While both stay on the straight path, R stays to the left of B. Moving into an empty neighbour never takes one through the other.”
CKSprout “What does that preserved order require of any successful route?”
Mina “One token must leave the path for A and later return. That needs an entry and an exit, in addition to their eight main-path moves.”
Main-path requirement
R needs at least 4 connections.
B needs at least 4 connections.
At least 8 moves
Necessary bay use
Someone enters A and later leaves.
These do not advance along the main path.
At least 2 more moves
Exclude a better result
Every successful route needs at least 8 + 2 = 10 moves.
Reach the bound legally
A legal ten-move sequence is still required.
Main-path steps cannot reverse the tokens’ order. A necessary bay entry and exit add two edges to the distance-only lower bound.
Without the bay, left-to-right order cannot change. The requested finish reverses that order, so a successful route must use the bay. The entry and exit make no progress along the main path and add at least two moves.
Bound, then build the best now has a stronger lower bound: at least 10 moves. It still needs a legal ten-move construction.
Act it out
Use two different buttons as tokens and sketch the empty map. Before looking at the route below, try to build a ten-move route yourself. Decide which token will wait, where it can wait, and when it can rejoin the path. Keep both positions visible; move one button only when its destination is empty. You may use either token as the waiting one.
Mina “When I move both buttons towards the middle, they block each other. I’ll move R into A before B reaches the junction.”
Rowan “Now B can pass. I’ll keep R there until B has cleared its next place, then bring R out. That gives me a complete route to count.”
CKSprout “Does every move in your model follow the original rule, and does the count reach our bound?”
Start
Initial positions
Move 1
R: 0 → 1
Move 2
R: 1 → 2
Move 3
R: 2 → A
Move 4
B: 4 → 3
Move 5
B: 3 → 2
Move 6
B: 2 → 1
Move 7
B: 1 → 0
Move 8
R: A → 2
Move 9
R: 2 → 3
Move 10
R: 3 → 4
Follow the numbered states. Only one token moves each time; R waits in A until B has passed. The tenth move reaches both required finishes.
Check: R uses six moves, including entry to and exit from A; B uses four. The complete state after every move is shown. A legal route proves possibility; the independent lower-bound argument proves that none can be shorter.
Revisit Find an invariant when allowed moves preserve a feature that constrains the finish.
A garden beyond its measurements
The outline is only part of their shared project. They also choose flowers with different, overlapping bloom times. Flowers available across the growing season can provide food for pollinators when an earlier group has finished blooming.
Mina and Rowan have settled particular mathematical goals: area and legal movement. They can now use those plans while making the garden welcoming in other ways.
Your turn
Think through these before opening the reasoning. Each has a complete answer and check; no writing is needed.
Many possible partners
All 15 volunteers split into two nonempty groups. Every person belongs to exactly one group. Each person in the first group meets each person in the second group once; there are no other meetings.
What is the largest total number of different meetings, and which group sizes achieve it?
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What is the largest total number of different meetings, and which group sizes achieve it?
Many possible partners
All 15 volunteers split into two nonempty groups. Every person belongs to exactly one group. Each person in the first group meets each person in the second group once; there are no other meetings.
Your answer
Enter the greatest meeting count and both group sizes, smaller first. Explain why no positive whole split does better.
These quantities fit. Your bound and the construction reaching it still need to be justified against every original condition.
Break it into steps
Step 1 of 5
If three people join one group, how many are in the other?
Fifteen minus three is twelve.
Step 2 of 5
How many cross-group meetings does that trial give?
Three times twelve is thirty-six. Compare it with more balanced allowed groups.
Step 3 of 5
What is the greatest meeting count?
Seven and eight give fifty-six, matching the whole-number upper bound.
Step 4 of 5
How many people are in the smaller group in the best split?
The smaller group has 7 people.
Step 5 of 5
How many are in the larger group in the best split?
The larger group has 8 people, giving 7×8=56 meetings.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the partners
Answer: The maximum is 56 meetings, with groups of 7 and 8.
If the group sizes are two positive whole numbers adding to 15, their product counts the meetings. The fixed-sum bound is 7.5² = 56.25. A whole meeting count cannot exceed 56. Groups of 7 and 8 achieve 7 × 8 = 56, include all 15 people and are both nonempty.
Exclude a better result
A whole meeting count is at most 56, from the bound 56.25.
Reach the bound legally
7 × 8 = 56 meetings, using all 15 people.
Rows represent seven people in one group and columns eight in the other. Each cell is one distinct cross-group meeting.
Remove the passing place
There are only three places in a straight chain, 0–1–2, with no side bay or other connection. R starts at 0 and B at 2; place 1 is empty. One move takes one token along one connection into an empty place. No jumps, sharing or simultaneous swaps are allowed.
Can the tokens exchange ends? If so, what is the minimum number of moves? If not, explain why.
CKSTEM Thinking CompassPlan and check your reasoning.
Can the tokens exchange ends? If so, what is the minimum number of moves? If not, explain why.
Remove the passing place
There are only three places in a straight chain, 0–1–2, with no side bay or other connection. R starts at 0 and B at 2; place 1 is empty. One move takes one token along one connection into an empty place. No jumps, sharing or simultaneous swaps are allowed.
Start
Requested finish
There is no bay or hidden connection. The start and requested finish are givens, not a claim that a route exists.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
Start
Requested finish
There is no bay or hidden connection. The start and requested finish are givens, not a claim that a route exists.Show the answer and reasoning: no bay
Answer:They cannot exchange ends. There is no finite minimum move count for this impossible task.
Every allowed move preserves the left-to-right order of the tokens. R starts left of B, but the requested finish puts it right of B. Neither token can step past the other into an empty neighbour. Counting the distance to the goal would give a lower bound, but would not supply a possible route.
Finish all the labels
There are 9 garden labels to paint. Two people are both available from the start. Every label needs exactly 5 uninterrupted minutes from one person. A person can paint only one label at a time, and a label cannot be split between people. There is no other preparation or switching time.
What is the earliest time by which all nine labels can be finished? Give a work allocation and a reason no earlier finish can work.
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What is the earliest time by which all nine labels can be finished? Give a work allocation and a reason no earlier finish can work.
Finish all the labels
There are 9 garden labels to paint. Two people are both available from the start. Every label needs exactly 5 uninterrupted minutes from one person. A person can paint only one label at a time, and a label cannot be split between people. There is no other preparation or switching time.
Your answer
Enter the earliest finish. Also supply an allocation and explain why an earlier finish cannot work.
The number fits. Your reasoning, legal construction and check of every rule remain unscored.
Break it into steps
Step 1 of 3
How many whole labels can one person finish in twenty minutes?
Twenty divided by five is four.
Step 2 of 3
How many labels could both finish by twenty minutes?
Together they finish at most eight, less than the required nine.
Step 3 of 3
What is the earliest finish for all nine labels?
Five and four labels finish by twenty-five minutes.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the labels
Answer: The minimum is 25 minutes. Give 5 labels to one person and 4 to the other.
If each person painted at most four labels, only eight would be finished. So someone must paint at least five whole labels, requiring 25 minutes. Allocating five and four lets them finish after 25 and 20 minutes, so all are ready after 25.
0 min5 min10 min15 min20 min25 min
Person 1
Label 1Label 2Label 3Label 4Label 5
Person 2
Label 6Label 7Label 8Label 9Free
Each cell is five minutes. One person paints five labels while the other paints four, so every job is finished at the twenty-five-minute mark.
Check: all nine jobs are assigned once, nobody paints two at the same time and no label is split. Dividing the total 45 minutes of work by two gives the weaker lower bound 22.5 minutes; the indivisible jobs prevent that bound being reached.
What makes “best” a proved result?
Name the goal and the allowed choices. Find a bound that covers every choice, then try to reach it legally. If your construction misses the bound, investigate: the construction may improve, the bound may be loose, or equality may be forbidden.
A finite list can supply a complete comparison. A preserved property can rule out a tempting shortcut. A model can uncover a workable construction. Whatever route you choose, return to both proof jobs and every original condition.
Mina “This outline fits our edging. Does that make it the best one?”
CKSprout “What does your outline establish, and what still needs a reason?”
Rowan “It shows one design is possible. We haven’t ruled out a bigger area.”