math-2017:G.MG.3
math-2017:G.MG.3 Apply geometric methods to solve design problems, e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios.
Learning goal: Apply geometric methods to solve design problems, such as designing an object or structure to satisfy physical constraints or to minimize cost, and working with typographic grid systems based on ratios, by expressing the design in terms of one variable, comparing cases systematically, and checking that every constraint is met.
What makes a problem a design problem. A design problem asks for the dimensions of something that does not exist yet. It has constraints, conditions the design must satisfy, such as a fixed volume, a maximum slope, a fixed length of fencing or a page size; and it usually has an objective, a quantity to make as small or as large as possible, such as material used, cost or enclosed area. Geometry supplies the formulas that connect the dimensions to the constraints and the objective. The method in this lesson has four steps: draw and label the object; use each constraint to write the objective in terms of a single variable; compare cases systematically, narrowing in on the best one; and verify that the chosen design meets every constraint, including the ones that were not part of the objective.
Minimizing material: an open box. A box with a square base of side s and height h, open at the top, must hold 4,000 cm³. The constraint is s²h = 4,000, so h = 4,000/s². The material is the base plus four sides: A = s² + 4sh = s² + 4s × 4,000/s² = s² + 16,000/s. Now compare cases. s = 10: h = 40, A = 100 + 1,600 = 1,700 cm². s = 15: h ≈ 17.78, A = 225 + 1,066.67 ≈ 1,291.67. s = 20: h = 10, A = 400 + 800 = 1,200. s = 25: h = 6.4, A = 625 + 640 = 1,265. s = 30: h ≈ 4.44, A = 900 + 533.33 ≈ 1,433.33. The area falls, bottoms out near s = 20 and rises again. Narrow the search: s = 18 gives 324 + 888.89 ≈ 1,212.89 and s = 22 gives 484 + 727.27 ≈ 1,211.27, both larger than 1,200. So the best design among these is s = 20 cm, h = 10 cm, using 1,200 cm² of material; the height is half the base side, not equal to it, because the open top means the base is the only horizontal panel. Comparing cases does not by itself prove that no value between 18 and 22 beats 20; it shows the minimum is close to 20. Algebra settles it exactly: A − 1,200 = s² + 16,000/s − 1,200 = (s − 20)²(s + 40)/s, which is positive for every s > 0 except s = 20, where it is 0. So 1,200 cm² at s = 20 is the true minimum, not just the best case tried.
Minimizing cost when materials differ. Suppose the base material costs 2 cents per cm² and the side material 1 cent per cm² (hypothetical prices). The cost is C = 2s² + 16,000/s cents. Cases: s = 12 gives 288 + 1,333.33 ≈ 1,621.33; s = 15 gives 450 + 1,066.67 ≈ 1,516.67; s = 16 gives 512 + 1,000 = 1,512; s = 17 gives 578 + 941.18 ≈ 1,519.18; s = 20 gives 800 + 800 = 1,600. The cheapest design is near s = 16 cm, with h = 4,000/256 ≈ 15.6 cm and cost about $15.12. The pricier base has pushed the design toward a smaller base and taller sides. This is the general lesson: the objective, not the shape alone, decides the best dimensions.
Maximizing area with a fixed perimeter. A rectangular field is to be fenced on three sides, the fourth side being a straight river, with 120 m of fencing (hypothetical). Let the two sides perpendicular to the river be x; the side parallel to it is 120 − 2x. Area A = x(120 − 2x). Cases: x = 20 gives 1,600; x = 25 gives 1,750; x = 30 gives 1,800; x = 35 gives 1,750; x = 40 gives 1,600. The table is symmetric about x = 30, and the algebra confirms it: x(120 − 2x) = 1,800 − 2(x − 30)², which is largest, 1,800 m², exactly when x = 30, for a field 30 m by 60 m. Note that the best rectangle is not a square: the river replaces one fence, so the design uses more length along the river.
Satisfying a physical constraint. Some design problems have no objective, only constraints that must all be met. A wheelchair ramp (hypothetical specification) may rise at most 1 unit for every 12 units of run. To reach a door 30 inches above the ground the run must be at least 30 × 12 = 360 inches, or 30 feet, and the ramp surface will be √(30² + 360²) ≈ 361.2 inches long. If the available space is only 25 feet, the constraint cannot be met by a straight ramp, and the design must change, for example by turning the ramp with a landing. Recognizing that a constraint is violated, and saying so, is part of the solution.
Typographic grid systems based on ratios. Page layout is a design problem where the constraints are ratios. A letter-size page is 8.5 by 11 inches; with 1-inch margins on every side the text area is 6.5 by 9 inches. Dividing the width into 3 columns with 0.25-inch gutters between them gives columns of (6.5 − 2 × 0.25) ÷ 3 = 2 inches. Type sizes in a modular scale are built from a ratio: starting at 12 points with ratio 1.25 the sizes are 12, 15, 18.75 and 23.44 points, each 1.25 times the last, so headings and body text relate by a constant factor. Paper in the A series has sides in the ratio √2 ≈ 1.414, so that cutting a sheet in half across its long side produces two sheets of the same proportions; A4 is 210 mm by 297 mm and 297 ÷ 210 ≈ 1.414. A text block designed with the ratio 1.618, the golden ratio, that is 10 cm wide is 16.18 cm tall. A grid built from ratios keeps every element proportional when the page is scaled, which is a direct application of similarity.
A worked model inside the lesson. A cylindrical can must hold 500 cm³ (about 500 mL) and use the least metal, counting the top, the bottom and the side. Constraint: πr²h = 500, so h = 500/(πr²). Surface area: A = 2πr² + 2πrh = 2πr² + 1,000/r. Cases: r = 3 gives 56.55 + 333.33 ≈ 389.9 cm²; r = 4 gives 100.53 + 250 ≈ 350.5; r = 4.3 gives 116.18 + 232.56 ≈ 348.7; r = 5 gives 157.08 + 200 ≈ 357.1; r = 6 gives 226.19 + 166.67 ≈ 392.9. The least material is near r = 4.3 cm, where h = 500 ÷ (π × 4.3²) ≈ 8.6 cm, about 348.7 cm². The height is close to twice the radius, so the best can is as tall as it is wide. Real cans are often taller than this, because the objective in practice includes handling, stacking and the cost of the seams, not just the area of metal.
Misconceptions, and why they are wrong. "The best box is always a cube" is false for an open box, where the best height is half the side. "Two cases are enough" is false; the area can be smaller between the two cases tried, so the comparison must bracket the best value and then narrow. "The objective is all that matters" ignores constraints; a can that uses the least metal but does not hold 500 cm³ is not a solution. Mixing units, such as cents and dollars or inches and feet, produces designs that are off by a constant factor. And reporting a design without checking it against every constraint is not finished; the check is the last step.
Worked open box by cases: An open-topped box with a square base must hold 4,000 cm³ and use the least material. With base side s, h = 4,000/s² and A = s² + 16,000/s. s = 10: 1,700 cm². s = 15: 1,291.67. s = 18: 1,212.89. s = 20: 1,200. s = 22: 1,211.27. s = 25: 1,265. s = 30: 1,433.33. The values fall to 1,200 at s = 20 and rise on both sides, so the design is a 20 cm by 20 cm base with height 10 cm, using 1,200 cm² of material. Check the constraint: 20 × 20 × 10 = 4,000 cm³.
Worked fenced field: With 120 m of fencing along three sides of a rectangle whose fourth side is a river, let x be a side perpendicular to the river. Area x(120 − 2x): 1,600 at x = 20, 1,750 at x = 25, 1,800 at x = 30, 1,750 at x = 35, 1,600 at x = 40. The maximum, 1,800 m², is at x = 30, giving a 30 m by 60 m field. Check: 30 + 30 + 60 = 120 m of fence.
Worked cost minimization: For the same open box, the base costs 2 cents per cm² and the sides 1 cent per cm². Cost C = 2s² + 16,000/s: 1,621.33 at s = 12; 1,516.67 at s = 15; 1,512 at s = 16; 1,519.18 at s = 17; 1,600 at s = 20. The cheapest of the cases is s = 16 cm, h = 4,000/256 = 15.625 cm, cost 1,512 cents = $15.12. The minimum-area design (s = 20) would cost 1,600 cents = $16.00, so the cheaper material for the sides changes the best shape.
Worked typographic grid: A page 8.5 in by 11 in has 1-inch margins, so the text area is 6.5 in by 9 in. Three columns with two 0.25-inch gutters: column width (6.5 − 0.5) ÷ 3 = 2 in. A modular type scale with ratio 1.25 from 12 pt gives 12, 15, 18.75 and 23.44 pt (rounded to 23.4). If the page is reproduced at 80% size, every measure scales by 0.8: columns 1.6 in, body type 9.6 pt, heading 18.75 pt, and the ratios between them are unchanged, which is the point of building the grid from ratios.
Design table record: For the 4,000 cm³ open box, make a table with rows s = 10, 15, 18, 20, 22, 25, 30 and columns h = 4,000/s², base area s², side area 4sh, total area, and cost at 2 cents per cm² for the base and 1 cent per cm² for the sides. The total area column should read 1,700, 1,291.67, 1,212.89, 1,200, 1,211.27, 1,265 and 1,433.33 cm². Mark the row with the least area and the row with the least cost and write one sentence explaining why they are different rows.
Independent performance: Pose and solve a design problem of your own with at least one constraint and one objective: a container of fixed volume using the least material, a fenced or walled region of fixed perimeter with the most area, a ramp or roof that must satisfy a slope limit, or a page layout with given margins, column count, gutter width and type-scale ratio. Success criteria: a labeled sketch names every dimension; the constraint is used to write the objective in one variable; at least five cases are compared in a table that brackets the best value and then narrows it; the chosen design is checked against every constraint with the arithmetic shown; the answer is reported with units and the correct precision; and one sentence states what the comparison of cases does and does not prove about the true minimum or maximum.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
An open-topped box with a square base must hold 4,000 cm³. Write the material area in terms of the base side s alone, and explain where the constraint was used.
The constraint s²h = 4,000 gives h = 4,000/s². The material is the base plus four sides: s² + 4sh = s² + 4s(4,000/s²) = s² + 16,000/s. The constraint was used to eliminate h, so the area depends on one variable and cases can be compared.
Using the cases s = 15, 18, 20, 22 and 25 for the open box, identify the best design and explain what the comparison does and does not prove.
Areas: 1,291.67, 1,212.89, 1,200, 1,211.27 and 1,265 cm². The smallest is 1,200 cm² at s = 20 cm, h = 10 cm. The pattern falls to s = 20 and rises on both sides, which shows the minimum is near 20 but cannot rule out a smaller value at, say, s = 19.5; more cases would still only be evidence. The proof is algebra: A − 1,200 = (s − 20)²(s + 40)/s, which is never negative for s > 0 and is 0 only at s = 20.
A rectangular field is fenced on three sides with 120 m of fencing, the fourth side being a river. Find the dimensions that enclose the most area.
With x the side perpendicular to the river, the parallel side is 120 − 2x and the area is x(120 − 2x): 1,600 at x = 20, 1,750 at 25, 1,800 at 30, 1,750 at 35, 1,600 at 40. The table suggests x = 30, and completing the square proves it: x(120 − 2x) = 1,800 − 2(x − 30)², which is at most 1,800 and equals 1,800 only at x = 30. The best field is 30 m by 60 m, area 1,800 m². Check the constraint: 30 + 30 + 60 = 120 m.
For the open box, the base costs 2 cents per cm² and the sides 1 cent per cm². Why does the cheapest design differ from the least-material design, and roughly what is it?
Cost is 2s² + 16,000/s, which weights the base twice as heavily as the material count does, so the best design shrinks the base and raises the sides. Cases give 1,516.67 cents at s = 15, 1,512 at s = 16 and 1,519.18 at s = 17, so the cheapest is near s = 16 cm with h ≈ 15.6 cm, about $15.12, against $16.00 for the s = 20 design.
A ramp may rise at most 1 unit for every 12 units of horizontal run and must reach a door 30 inches above the ground. Find the minimum run and the ramp length, and say what happens if only 25 feet of run is available.
Run ≥ 30 × 12 = 360 inches = 30 feet. Ramp length √(30² + 360²) ≈ 361.2 inches, about 30.1 feet. With only 25 feet = 300 inches of run the slope would be 30/300 = 1/10, steeper than 1/12, so a straight ramp violates the constraint; the design must change, for example by adding a turn with a landing.
A letter-size page (8.5 by 11 in) has 1-inch margins and three text columns separated by 0.25-inch gutters. Find the column width.
Text width 8.5 − 2 = 6.5 in. Two gutters take 0.5 in, leaving 6.0 in for three columns, so each column is 2 in wide. The text height is 11 − 2 = 9 in and does not affect the column width.
A modular type scale starts at 12 pt with ratio 1.25. List the next three sizes, and explain what happens to the ratios if the whole page is printed at 80% size.
12 × 1.25 = 15 pt, 15 × 1.25 = 18.75 pt, 18.75 × 1.25 ≈ 23.44 pt. At 80% every size is multiplied by 0.8 (9.6, 12, 15, 18.75 pt), and the ratio between consecutive sizes is still 1.25 because both sizes were scaled by the same factor; a grid built from ratios survives scaling.
A can must hold 500 cm³ and use the least metal for its top, bottom and side. Write the surface area in terms of r and compare r = 3, 4, 4.3 and 5 cm.
h = 500/(πr²), so A = 2πr² + 2πrh = 2πr² + 1,000/r. r = 3: 389.9 cm²; r = 4: 350.5; r = 4.3: 348.7; r = 5: 357.1. The least metal is near r = 4.3 cm, where h ≈ 8.6 cm, about twice the radius. Every case satisfies the volume constraint because h was chosen from it.