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Designing Under Constraints: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Check three sign proposals

A notice board fits a rectangular sign at most $115$ cm wide and $170$ cm high. The club has at most $16{,}000$ square centimeters of usable card. Treat the sign as a single rectangle and assume no card is lost in cutting.

Plan A is $96$ cm wide and $155$ cm high.

Plan B is $110$ cm wide and $149$ cm high.

Plan C is $120$ cm wide and $126$ cm high.

Which plans are feasible? If the club wants the most display area among these plans, which should it choose?
1. Hint 1
  Check each limit separately before ranking the plans.
2. Hint 2
  Find the area of each plan and compare its width, height, and area with their respective caps.
Answer

Only A is feasible. Choose A for its $14{,}880$ square centimeters of display area.

Full solution

Plan A fits both dimension caps.
Its area is $96(155)$ square centimeters, or $14{,}880$ square centimeters, within the card limit.

Plan B fits the dimension caps, but its area is $110(149)$ square centimeters, or $16{,}390$ square centimeters, above the card limit.

Plan C uses $120(126)$ square centimeters of card, or $15{,}120$ square centimeters, but it is wider than the space permits.

A is the only feasible proposal, so it gives the most display area among these choices.

Answer

Only A is feasible. Choose A for its $14{,}880$ square centimeters of display area.

Key idea

A plan must meet every cap before its objective value can earn it a place in the comparison.
Problem 2 Make room for a gate

A rectangular outdoor waiting area is $11$ m by $7$ m. It needs a $2$ m gate on one of the $11$ m sides and new fencing on the rest of the perimeter. Fencing costs $8$ dollars per meter, and the gate costs $39$ dollars. Assume the only costs are the gate and the exact length of fence used.

Find the total cost. If the budget is $320$ dollars, say whether the design is feasible and give the difference between cost and budget.

The gate fills the gap in one long side.

Text description of this figure

An 11 meter by 7 meter rectangular enclosure. A 2 meter gap centered in the top 11 meter side holds the gate.
1. Hint 1
  The gate and fencing cover different parts of the boundary.
2. Hint 2
  Find the perimeter, then take out the length supplied by the gate before pricing the fence.
Answer

The total is $311$ dollars. The design is feasible with $9$ dollars left.

Full solution

The boundary length is $2(11+7)$ meters, or $36$ meters.
The gate fills $2$ m of that boundary, leaving $34$ m for fence.

Pay for that fencing and the gate separately.
$34(8)+39=311.$

The total is within budget.
The difference between $320$ dollars and $311$ dollars is $9$ dollars.

Answer

The total is $311$ dollars. The design is feasible with $9$ dollars left.

Key idea

Charge each part of the perimeter at the rate that applies to it.
Problem 3 Clearance between pots

A shelf is $74$ cm long. Each flowerpot takes $11$ cm of shelf length. Neighboring pots must have a gap of at least $3$ cm between them, but there is no required gap at either end of the shelf.

Find the largest number of pots in one row. How much shelf length is left over when all gaps in that row are exactly $3$ cm?

Sample of the spacing rule. Use the full shelf length given in the problem.

Text description of this figure

A sample row with three pots. Each pot uses 11 centimeters of shelf width. The gaps between neighboring pots are 3 centimeters. No gap is required at the ends.
1. Hint 1
  Gaps occur between neighbors, so count pots and gaps separately.
2. Hint 2
  Compare the length used by several consecutive pot counts, with one fewer gap than pots.
Answer

At most $5$ pots fit, with $7$ cm left at minimum spacing.

Full solution

Five pots have four gaps between them.
$5(11)+4(3)=67.$

Six pots would have five gaps.
$6(11)+5(3)=81.$

Six pots exceed the shelf length even with minimum gaps.
Adding more pots only increases the required length.
Five fit and leave $74-67$ cm, or $7$ cm.

Answer

At most $5$ pots fit, with $7$ cm left at minimum spacing.

Key idea

Use the minimum gap required between neighbors when testing how many items can fit.
Problem 4 Pay for whole sheets

A theater needs $30$ case liners. It can cut them from stock sheets or order them already cut.

Stock-sheet order. Each liner needs $0.26$ square meters. The cutting layout requires another $0.80$ square meters of total waste for the whole job. Each sheet provides $4$ square meters and costs $15$ dollars. The cutter has confirmed the job fits on any order with enough total area, and sheets are sold whole.

Precut order. Each liner costs $1.42$ dollars, including all charges.

Which order costs less and by how much?
1. Hint 1
  Decide what you actually pay for in each order.
2. Hint 2
  For the stock order, add waste to the total liner area and round the number of sheets up before pricing.
Answer

The precut order costs $42.60$ dollars and is $2.40$ dollars cheaper than the $45$ dollar stock-sheet order.

Full solution

Stock-sheet order.
The job uses $30(0.26)+0.80$ square meters, or $8.60$ square meters.
Two sheets provide only $8$ square meters.
Three sheets provide $12$ square meters and cost $45$ dollars.

Precut order.
Multiply the price per liner by the quantity.
$30(1.42)=42.60.$

The precut order saves $45-42.60$ dollars, or $2.40$ dollars.

Answer

The precut order costs $42.60$ dollars and is $2.40$ dollars cheaper than the $45$ dollar stock-sheet order.

Key idea

A sheet purchase counts whole sheets even if some of the last sheet will be left over.
Problem 5 More storage with the same coating supply

A craft studio can choose one of three closed display boxes. It can coat at most $550\text{ cm}^2$ and wants the largest inside volume. The coating covers all six exterior faces. Treat each board as having negligible thickness so one set of dimensions serves for both volume and face area.

Box A is $14$ cm by $9$ cm by $6$ cm.

Box B is $12$ cm by $12$ cm by $5$ cm.

Box C is $15$ cm by $10$ cm by $6$ cm.

For each box, find the area to coat and the inside volume. Choose the feasible box that meets the objective best.
1. Hint 1
  Capacity and coating describe different geometric quantities.
2. Hint 2
  Use two copies of each face area for coating, then multiply the three dimensions for volume.
Answer

Box A is best.
$\begin{aligned}S_A&=528\text{ cm}^2,\\V_A&=756\text{ cm}^3,\\S_B&=528\text{ cm}^2,\\V_B&=720\text{ cm}^3,\\S_C&=600\text{ cm}^2,\\V_C&=900\text{ cm}^3.\end{aligned}$

Full solution

Box A.
Each of its three distinct face positions contributes a pair of faces.
$S_A=2(126+84+54).$
The area is $528$ square centimeters.
Multiplying $14$ , $9$ , and $6$ gives $756$ cubic centimeters.

Box B.
Its face pairs give $2(144+60+60)$ square centimeters, or $528$ square centimeters.
Multiplying $12$ , $12$ , and $5$ gives $720$ cubic centimeters.

Box C.
Its face pairs give $2(150+90+60)$ square centimeters, or $600$ square centimeters.
Multiplying $15$ , $10$ , and $6$ gives $900$ cubic centimeters.

The coating cap rules out C.
Between A and B, A provides the larger volume.

Answer

Box A is best.
$\begin{aligned}S_A&=528\text{ cm}^2,\\V_A&=756\text{ cm}^3,\\S_B&=528\text{ cm}^2,\\V_B&=720\text{ cm}^3,\\S_C&=600\text{ cm}^2,\\V_C&=900\text{ cm}^3.\end{aligned}$

Key idea

Surface area can decide feasibility while volume decides which feasible design wins.
Problem 6 Challenge a partial table

A music studio can install a rectangular acoustic mat. The side parallel to the window must be at most $4$ m and the perpendicular side at most $8$ m. Both lengths must be positive whole meters. Edging costs $6$ dollars per meter on each parallel side and $2$ dollars per meter on each perpendicular side. The budget is $56$ dollars. The objective is to maximize mat area.

Tested layout A is $2$ m parallel by $6$ m perpendicular, with area $12$ square meters.

Tested layout B is $3$ m parallel by $4$ m perpendicular, with area $12$ square meters.

Tested layout C is $4$ m parallel by $2$ m perpendicular, with area $8$ square meters.

A studio member says the table proves a maximum of $12$ square meters. Give one feasible counterexample and explain what it proves.
1. Hint 1
  A single untested feasible layout that improves the area is enough to refute the claim.
2. Hint 2
  For one permitted parallel length, work out how much money remains to edge the two perpendicular sides.
Answer

One counterexample is $2$ m parallel by $8$ m perpendicular. It costs $56$ dollars and covers $16$ square meters, so $12$ square meters is not the maximum.

Full solution

Layout A fits both dimension caps and costs $2(2)(6)+2(6)(2)$ dollars, or $48$ dollars.
It has $8$ dollars of budget left.

Keep A's parallel length and extend each perpendicular side by $2$ m.
The $4$ extra meters of edging cost $8$ dollars.

The revised layout is $2$ m by $8$ m, within both dimension caps.
Its area is $16$ square meters and its cost is $56$ dollars.

That is one feasible design with area above the tested best.
A $3$ m by $5$ m layout also works, with cost $56$ dollars and area $15$ square meters.

Answer

One counterexample is $2$ m parallel by $8$ m perpendicular. It costs $56$ dollars and covers $16$ square meters, so $12$ square meters is not the maximum.

Key idea

A counterexample can disprove an optimum without finding the true optimum.
Problem 7 Leave room for a clear strip

Two stage layouts both fit the site. Each must leave one clear rectangular strip with the same prescribed width $w$ , measured in meters. Performers get all the remaining floor area. Either layout is feasible for any positive width up to $2$ m.

Plan A is $8$ m by $6$ m. Its strip runs along the full $8$ m side.

Plan B is $9$ m by $5$ m. Its strip runs along the full $5$ m side.

Case 1. Which plan has more performance area at strip width $0.8$ m?

Case 2. Which plan has more performance area at strip width $1.2$ m?

At what width do the plans tie?

Shaded strips have the same width. Their lengths differ.

Text description of this figure

Plan A is 8 meters horizontally and 6 meters vertically, with a horizontal clear strip along the 8 meter side. Plan B is 9 meters horizontally and 5 meters vertically, with a vertical clear strip along the 5 meter side.
1. Hint 1
  The longer clear strip removes more area per meter of width.
2. Hint 2
  Write each performance area as the whole rectangle area minus the strip area. Set the two expressions equal to find the tie.
Answer

Case 1. A has $41.6$ square meters versus B's $41$ . Case 2. B has $39$ square meters versus A's $38.4$ . The plans tie at a strip width of $1$ m.

Full solution

For A, subtract an $8$ m long strip.
The performance area in square meters is $48-8w$ .

For B, subtract a $5$ m long strip.
The performance area in square meters is $45-5w$ .

Case 1.
At $w=0.8$ , the areas are $41.6$ square meters for A and $41$ square meters for B.
A wins.

Case 2.
At $w=1.2$ , the areas are $38.4$ square meters for A and $39$ square meters for B.
B wins.

A starts with $3$ more square meters, but its strip removes $3$ more square meters for each meter of strip width.
The areas tie when $3w=3$ , at $w=1$ m.
Both have $40$ square meters then.

Answer

Case 1. A has $41.6$ square meters versus B's $41$ . Case 2. B has $39$ square meters versus A's $38.4$ . The plans tie at a strip width of $1$ m.

Key idea

The length of a reserved strip determines how quickly its widening reduces usable area.
Problem 8 What the row count leaves open

A shelf is $28$ cm wide and $21$ cm deep. Its rectangular trays are $8$ cm by $5$ cm, and they can touch without overlapping. An exhaustive check of parallel full-width rows, with one orientation per row, finds a maximum of $13$ trays.

First verify that $13$ trays actually fit using two rows that are each $8$ cm deep and one row that is $5$ cm deep.

The owner now permits any placement of the trays in a single layer without overlaps. Use the areas of the shelf and tray to give an upper bound for the number that fit.

Find the true maximum under the new rule and justify it.
1. Hint 1
  Separate an arrangement that fits from an upper bound based on area.
2. Hint 2
  Divide the shelf area by the area of one tray. Ask whether the next whole number can fit by area alone.
3. Hint 3
  Try splitting the shelf into two side-by-side regions that use different layouts.
Answer

The full-width row layout fits 13. The area bound is 14. An 8 cm column holds 4 trays, and three rows in the remaining width hold 4, 4, and 2, so the true maximum is 14.

Full solution

An $8$ cm deep row fits five trays across, each $5$ cm wide.
A $5$ cm deep row fits three trays across, each $8$ cm wide.

The two deep rows and one shallow row use $21$ cm of depth and fit $5+5+3$ trays, or $13$ .
This arrangement is still allowed after the rule changes.

The shelf area is $588$ square centimeters and each tray covers $40$ square centimeters.
Fifteen trays cover $600$ square centimeters, so they cannot fit.

Fourteen trays need $560$ square centimeters.
That passes the area test, which by itself cannot certify a placement.
The true maximum is at least $13$ and at most $14$ .
Showing a placement of $14$ or ruling out all such placements would close the gap.

One placement closes the gap.
In a column $8$ cm wide, four trays lie across the column and use $20$ cm of depth.

In the remaining $20$ cm of width, two $8$ cm deep rows hold four trays each.
A $5$ cm deep row holds two more.
This side uses the full $21$ cm of depth.

The placement fits $4+8+2=14$ trays.
The area bound now proves that $14$ is the maximum.

The left column holds four trays. The right side holds four, four, and two.

Answer

The full-width row layout fits 13. The area bound is 14. An 8 cm column holds 4 trays, and three rows in the remaining width hold 4, 4, and 2, so the true maximum is 14.

Key idea

An upper bound and an arrangement attaining it prove a maximum.
Problem 9 A stencil inside a triangular banner

A right triangular banner has a horizontal leg of $18$ dm and a vertical leg of $12$ dm. A rectangular stencil shares the banner's right-angle corner and has its sides along those two legs. The corner opposite the shared corner lies on the hypotenuse. Both stencil dimensions may be any positive real lengths satisfying these conditions.

Find the stencil dimensions with the largest area. Prove that your area is greatest over all permitted widths.

One possible stencil. Its width and height can vary.

Text description of this figure

The banner is a right triangle with an 18 decimeter horizontal leg and a 12 decimeter vertical leg. The rectangular stencil shares the lower-left right-angle corner. Its upper-right corner lies on the sloping edge. Its width w and height h are variable.
1. Hint 1
  The sloping edge makes the allowed height fall as the width grows.
2. Hint 2
  Use similarity of the triangle to the right of the stencil and the whole banner to express the height in terms of the width.
3. Hint 3
  Once you have an area expression, look for a squared difference that gives its distance below an upper bound.
Answer

Width $9$ dm and height $6$ dm give the greatest area, $54\text{ dm}^2$ .

Full solution

Let $w$ and $h$ be the width and height in decimeters.
To the right of the stencil is a right triangle with base $18-w$ and height $h$ .
Its acute angle at the banner's lower-right vertex is shared with the banner, so the two triangles are similar.
$\frac{h}{18-w}=\frac{12}{18}.$

The height is $\tfrac23(18-w)$ , so the area is
$A=\tfrac23w(18-w).$
The feasible widths satisfy $0<w<18$ .
Every such width gives a positive height.

The two factors $w$ and $18-w$ sum to $18$ .
Rewrite their product using $w=9+(w-9)$ and $18-w=9-(w-9)$ .
$w(18-w)=81-(w-9)^2.$

The square is nonnegative, so the product is at most $81$ .
Multiplying by two-thirds gives an area of at most $54$ square decimeters.

At width $9$ dm, the square vanishes.
The height is $6$ dm and the area reaches the bound.
Width $9$ dm belongs to the feasible interval, so this proves the maximum.

Answer

Width $9$ dm and height $6$ dm give the greatest area, $54\text{ dm}^2$ .

Key idea

An upper bound plus a feasible design that reaches it proves a maximum over the modeled dimensions.
Problem 10 Move an order to a shallower shelf

A repair shop must choose one of three available closed boxes. The camera kit needs at least $1200\text{ cm}^3$ of inside capacity. Each box is loaded in the stated orientation, with the first dimension along the shelf front, the second running front to back, and the third vertical. The old shelf allows up to $21$ cm along the front, $15$ cm front to back, and $6$ cm in height.

Box A is $20$ cm by $12$ cm by $5$ cm.

Box B is $16$ cm by $15$ cm by $5$ cm.

Box C is $18$ cm by $12$ cm by $6$ cm.

Ignore board thickness and waste. All board has the same cost per square centimeter, and there are no fees.

Old shelf. Which feasible box costs least? Justify your choice.

New shelf. The front-to-back clearance falls to $14$ cm. With the same capacity requirement and other clearances, which feasible box now costs least? Justify your revised choice.
1. Hint 1
  Check which box can physically fit each shelf before you compare its cost.
2. Hint 2
  Because the board rate is the same for all boxes, compare the areas of the six faces of the feasible boxes.
Answer

Old shelf. Choose B, which uses $790$ square centimeters of board. New shelf. Choose C, which uses $792$ square centimeters of board.

Full solution

Box A.
The inside volume is $20(12)(5)$ cubic centimeters, or $1200$ .
The board area is $2(240+100+60)$ square centimeters, or $800$ .

Box B.
The inside volume is $16(15)(5)$ cubic centimeters, or $1200$ .
The board area is $2(240+80+75)$ square centimeters, or $790$ .

Box C.
The inside volume is $18(12)(6)$ cubic centimeters, or $1296$ .
The board area is $2(216+108+72)$ square centimeters, or $792$ .

Old shelf.
All three boxes meet the capacity and clearance limits.
B uses the least board and has the lowest cost because the rate is the same.

New shelf.
B is $15$ cm front to back and exceeds the new clearance.
A and C still fit and meet the capacity requirement.
C uses $8$ fewer square centimeters of board than A, so C is least costly among all feasible available boxes.

Answer

Old shelf. Choose B, which uses $790$ square centimeters of board. New shelf. Choose C, which uses $792$ square centimeters of board.

Key idea

A revised clearance can rule out the former winner even when the costs stay the same.

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