Geometric Modeling and Design: Chapter Test
20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.
Multiple choice
20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.
Core practice
10 problems from across the chapter. 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)
0 of 10 done · 0 skipped · 0 to revisit
Progress saved in this browser.
Progress can't be saved in this browser, so your choices last for this visit only.
-
Problem 1 Two gauges at the same opening
Two gauges measure the same exact width of a machine opening. Gauge A reports 48 mm to the nearest 0.6 mm. Gauge B reports 48.4 mm to the nearest 1.1 mm. The only source of uncertainty is rounding. Both gauges round exact halfway readings to the larger report.
Find the full range of widths that agree with both reports. Explain which endpoints can occur.
- Hint 1
Each report admits widths up to half of its own reporting step away.
- Hint 2
Find the range for each gauge. A width that agrees with both reports must belong to both ranges.
Answer
The possible widths , in millimeters, are
A width of 47.85 mm can occur. A width of 48.3 mm cannot.
Full solution
Gauge A's half-step is 0.3 mm.
The lower halfway value rounds to 48, and the upper one rounds to 48.6.
Its range is
Gauge B's half-step is 0.55 mm.
Its lower halfway value rounds to 48.4, and its upper one rounds to 49.5.
Its range is
The higher lower bound and the lower upper bound give the overlap.
The lower endpoint is included by B's tie rule and lies inside A's interval.
The upper endpoint is excluded by A's tie rule.
Filled dots include their endpoint. Open dots exclude theirs. Answer
The possible widths , in millimeters, are
A width of 47.85 mm can occur. A width of 48.3 mm cannot.
Key idea
When reports concern the same measurement, keep only the readings allowed by every report.
- Hint 1
-
Problem 2 Brine in a stepped container
The inside of a container consists of two connected rectangular sections stacked vertically. From the floor to 6.5 cm above it, the horizontal cross-sectional area is 84 square centimeters. From that level to the rim, the cross-sectional area is 60 square centimeters and the section is 4.5 cm high.
The container already holds 0.72 kg of brine at density 1.25 grams per cubic centimeter. Assume uniform density, a level liquid surface, and no trapped air.
Find the current depth of the brine above the floor. How many more kilograms of the same brine are needed to fill the container to the rim?
The area labels refer to horizontal slices. The height labels are vertical. Text description of this figure
The lower rectangular section is 6.5 centimeters high, with horizontal cross-sectional area 84 square centimeters. The upper rectangular section is 4.5 centimeters high, with horizontal cross-sectional area 60 square centimeters. No liquid level is marked. The drawing is schematic.
- Hint 1
Compare the volume already in the container with the volume of the lower section.
- Hint 2
Once you know which section contains the surface, use that section's cross-sectional area to find how far into it the liquid extends.
- Hint 3
Find the total brine mass at the rim and account for the brine that is already there.
Answer
The brine is 7 cm deep. It takes another 0.30 kg of brine to reach the rim.
Full solution
The lower section's capacity, in cubic centimeters, is
The 0.72 kg of brine is 720 g.
The current volume in cubic centimeters is
The brine fills the lower section and occupies another 30 cubic centimeters in the upper section.
This adds a depth, in centimeters, of
The total depth is 7 cm.
The upper section contributes another 270 cubic centimeters at the rim.
The full volume is
The mass of brine at the rim in grams is
That is 1.02 kg.
The additional mass is
Answer
The brine is 7 cm deep. It takes another 0.30 kg of brine to reach the rim.
Key idea
At a step in a container, use the cross-sectional area of the layer where the liquid level changes.
- Hint 1
-
Problem 3 Counting burrows in a reserve
A reserve consists of two nonoverlapping zones. The ridge covers 0.7 square kilometers. The plain covers 1.8 square kilometers. A 0.1-square-kilometer sample on the ridge contains 19 burrows. A sample of the same area on the plain contains 11 burrows.
Estimate A. Use the two samples together to predict one density for the whole reserve. Find its predicted total burrow count.
Estimate B. Use the ridge sample for the ridge and the plain sample for the plain. Find the predicted total for the reserve.
An independent full count finds 340 burrows. Which estimate is closer? Explain why the estimates differ and whether the full count establishes the accuracy of the predicted count in either zone.
- Hint 1
The pooled samples give the ridge and plain the same amount of sampled area. How do the zone areas themselves compare?
- Hint 2
For the separate-zone estimate, convert each sample to a rate per square kilometer and apply that rate to the corresponding zone.
- Hint 3
Compare the prediction errors for the two totals. Ask whether the independent count tells you how its burrows are divided between zones.
Answer
Estimate A predicts 375 burrows. Estimate B predicts 331 and is closer, with errors of 35 and 9. The estimates differ because the zones have unequal areas despite equal sample areas. The reserve total cannot verify either zone count.
Full solution
Estimate A.
The pooled sample covers 0.2 square kilometers and contains 30 burrows.
It gives an estimated density of
Estimate A.
The whole reserve covers 2.5 square kilometers.
The predicted count is
Estimate B.
The ridge sample represents 190 burrows per square kilometer.
Apply that rate to the ridge to get
Estimate B.
The plain sample represents 110 burrows per square kilometer.
Apply that rate to the plain to get
Estimate B.
The two predicted counts add to
The pooled density gives the two sample rates equal weight.
The plain covers more of the reserve and has the lower estimated density, so using the zone areas makes the prediction lower.
Estimate A is 35 above the full count.
Estimate B is 9 below it, so B is closer.
The full count gives no division into zones.
Overestimates and underestimates for individual zones could offset each other.
Answer
Estimate A predicts 375 burrows. Estimate B predicts 331 and is closer, with errors of 35 and 9. The estimates differ because the zones have unequal areas despite equal sample areas. The reserve total cannot verify either zone count.
Key idea
When samples represent different-sized regions, weighting by regional area can change the total prediction.
- Hint 1
-
Problem 4 The missing scale on a habitat plan
A printed plan uses the same length scale in both directions, but its scale label is missing. An outline for Zone R covers 36 square centimeters on the plan. A full count found 72 trees in Zone R, and a field report gives its exact average density as 1,250 trees per square kilometer.
Recover the actual length in meters represented by 1 cm on the plan.
Zone S covers 27 square centimeters on the same plan and contains 81 trees. Find its actual average tree density in trees per square kilometer.
- Hint 1
The count and the average density determine the actual area of Zone R.
- Hint 2
Find how many square meters correspond to one square centimeter on the plan. Imagine that one square centimeter as a square.
- Hint 3
Use the recovered scale to find Zone S's actual area before calculating its density.
Answer
One centimeter on the plan represents 40 m. Zone S averages 1,875 trees per square kilometer.
Full solution
Zone R's actual area in square kilometers is
That is 57,600 square meters.
Each square centimeter of the plan represents this many square meters.
A plan square 1 cm on a side represents an actual square.
Its side length in meters is
Zone S has actual area
That is 0.0432 square kilometers.
Its average in trees per square kilometer is
Answer
One centimeter on the plan represents 40 m. Zone S averages 1,875 trees per square kilometer.
Key idea
A known area can recover the length scale, which then applies to other regions drawn on the same plan.
- Hint 1
-
Problem 5 An aisle through the tray bay
A storage bay is 23 cm deep. A 5 cm wide aisle runs the full depth and separates two rectangular storage zones. Each tray is 6 cm by 5 cm. Each zone uses one layer in a rectangular grid, with tray edges parallel to the zone edges and all trays within that zone facing the same way. The zones can use different orientations. Trays can touch, and no tray may extend into the aisle.
Current aisle. The zone widths are 14 cm on the left and 27 cm on the right. Find the largest total tray count and justify it.
Moved aisle. The zone widths become 17 cm on the left and 24 cm on the right. Find the largest total tray count and justify whether moving the aisle increases it.
Schematic plan views. The shaded aisle is unavailable for storage. Text description of this figure
Both plan views have depth 23 centimeters. In the current position, a 5-centimeter aisle separates the left zone of width 14 centimeters from the right zone of width 27 centimeters. In the moved position, the left zone has width 17 centimeters and the right zone has width 24 centimeters. No trays are shown.
- Hint 1
The aisle keeps the two counts separate. For each zone there are only two ways to orient the trays.
- Hint 2
For each orientation, count how many whole tray widths fit across and how many whole tray depths fit from front to back.
- Hint 3
Take the better count in each zone before combining the zones. Repeat that comparison after the aisle moves.
Answer
The current aisle allows at most 24 trays, with 8 on the left and 16 on the right. The moved aisle allows at most 25 trays, with 9 on the left and 16 on the right. Moving the aisle adds one tray.
Full solution
A tray with 6 cm across uses 5 cm of depth, so up to four rows fit.
A tray with 5 cm across uses 6 cm of depth, so up to three rows fit.
These are the only two orientations for each grid.
Current aisle, left zone.
With 6 cm across, two columns fit.
With 5 cm across, two columns fit.
The counts are
Current aisle, right zone.
With 6 cm across, four columns fit.
With 5 cm across, five columns fit.
The counts are
Current aisle.
The two best grids fit at the same time because their zones are disjoint.
The maximum is
Moved aisle, left zone.
With 6 cm across, two columns fit.
With 5 cm across, three columns fit.
The counts are
Moved aisle, right zone.
With 6 cm across, four columns fit.
With 5 cm across, four columns fit.
The counts are
Moved aisle.
The maximum is
The move adds one tray.
The counts are exact maxima because each orientation was checked in each zone and the two best grids can be installed together.
Answer
The current aisle allows at most 24 trays, with 8 on the left and 16 on the right. The moved aisle allows at most 25 trays, with 9 on the left and 16 on the right. Moving the aisle adds one tray.
Key idea
An aisle can change the best orientation in one zone even when the total storage width stays fixed.
- Hint 1
-
Problem 6 A trench measured two ways
A trench is modeled as a 12.5 m long prism of constant cross section. Its top width is fixed at 1.6 m. Its depth is reported as 0.8 m to the nearest 0.2 m, with exact halfway readings going to the larger report. Treat the listed fixed dimensions as exact and track only rounding of the depth.
Model R. The cross section is a rectangle.
Model T. The cross section is a trapezoid with horizontal top and bottom edges and a fixed bottom width of 1.2 m.
Find the exact prediction interval for the volume under each model. An independent survey first gives a volume from 15.7 to 15.9 cubic meters, inclusive. A later, more precise survey gives 15.75 to 15.85 cubic meters, inclusive. Which models are consistent with each survey?
End views. Both models use the same reported depth. Text description of this figure
The rectangular cross section has top width 1.6 meters. The trapezoidal cross section has top width 1.6 meters and bottom width 1.2 meters. Their vertical depths are labeled h. The outlines are schematic.
- Hint 1
The two models share the same range of possible depths.
- Hint 2
For each cross section, write its area in terms of depth and multiply by the fixed trench length.
- Hint 3
An interval can touch a survey bound without sharing a value. Check whether the touching endpoint belongs to both intervals.
Answer
Model R predicts
Model T predicts
Volumes are in cubic meters. Both models agree with some volume from the first survey. Only Model R agrees with the more precise survey.
Full solution
The rounding step is 0.2 m, so the depth in meters satisfies
Model R.
The rectangle area is .
Multiplying by 12.5 gives
The volume range is
Model T.
The mean of the top and bottom widths is 1.4 m.
Multiplying this width by the depth and trench length gives
The volume range is
First survey.
Its range is inside R's prediction.
It also includes 15.7, which T allows.
Both models are consistent with this survey.
More precise survey.
Its range is inside R's prediction.
The survey starts at 15.75, and T allows only volumes strictly below 15.75.
Only R is consistent with the more precise survey.
Only the first survey includes values strictly below the cutoff. Answer
Model R predicts
Model T predicts
Volumes are in cubic meters. Both models agree with some volume from the first survey. Only Model R agrees with the more precise survey.
Key idea
Compare prediction and observation intervals using their included values, especially when two bounds touch.
- Hint 1
-
Problem 7 A tank on a lift
A prototype tank has 260 cubic centimeters of inside capacity and contains 35 cubic centimeters of wall and floor material. For each full-size design, all corresponding inside and outside lengths are multiplied by the same factor . Any real factor of at least 1 is allowed. The wall and floor material has density 2.6 grams per cubic centimeter.
The full-size tank is lifted when filled to its rim with liquid of density 0.85 grams per cubic centimeter. The lift supports at most 39 kg total, including the tank and its contents.
Find the largest possible scale factor. At delivery, the pump leaves 2.2 L in a tank that contains at least that much. It delivers nothing from a tank that holds less. Can any size that the lift can carry deliver at least 31 L in one trip? Justify your answer.
- Hint 1
The same length scale changes the volume of the tank material and the volume of liquid held to the rim.
- Hint 2
Calculate the masses for the prototype volumes, add them, and scale the combined mass.
- Hint 3
The mass limit gives the largest possible inside volume. Compare what can be pumped from that volume with the delivery goal.
Answer
The largest scale factor is 5. No size the lift can carry can deliver 31 L in one trip. The largest size holds 32.5 L and can deliver 30.3 L.
Full solution
For the prototype wall and floor volume, the material mass in grams is
For the prototype inside volume, the liquid mass in grams is
Both volumes scale by , so the combined mass in grams is
The limit is 39,000 g.
Dividing it by the mass coefficient gives
Since is positive, the largest permitted factor is 5.
It reaches the lift limit exactly.
At this scale the inside volume is
That is 32.5 L.
After allowing for the unpumpable liquid, the delivery in liters is
Any smaller scale has less inside capacity, so no size can deliver 31 L.
Answer
The largest scale factor is 5. No size the lift can carry can deliver 31 L in one trip. The largest size holds 32.5 L and can deliver 30.3 L.
Key idea
When the wall and interior scale together, the full load scales by the same volume factor as the capacity.
- Hint 1
-
Problem 8 Ordering plugs for a divided bed
A planting bed of exact area 137 square meters is divided into sunny and shaded parts with no overlap. The sunny area is reported as 51 square meters to the nearest 3 square meters, with halfway readings going to the larger report. Track only rounding uncertainty.
The crew budgets 4 plugs per square meter of sunny area and 7 plugs per square meter of shaded area. It adds the two allowances and rounds the sum up to a whole number of plugs. The supplier sells packs of 27 plugs.
Find the least number of packs that guarantees enough plugs for every possible sunny area. Give one possible sunny area for which one fewer pack would be enough.
- Hint 1
A larger sunny area leaves a smaller shaded area. Express both areas using a single variable.
- Hint 2
Decide which end of the sunny-area interval gives the largest total allowance. Check whether that end can occur.
- Hint 3
Compare the greatest whole-plug requirement with the number in 30 packs. Then try the reported sunny area itself.
Answer
The crew needs 31 packs to guarantee enough. If the sunny area is 51 square meters, 30 packs are enough.
Full solution
If the sunny area is in square meters, rounding gives
The shaded area is .
The plug allowance before rounding is
Simplifying gives
The allowance is largest when the sunny area is smallest.
The included endpoint gives
The crew rounds this allowance up to 811 plugs.
Thirty packs give
This falls short for .
Thirty-one packs give 837 plugs, enough for every possible area.
For the allowed sunny area , the allowance is
No rounding up is needed, and 30 packs supply 810 plugs.
Answer
The crew needs 31 packs to guarantee enough. If the sunny area is 51 square meters, 30 packs are enough.
Key idea
When two parts share a fixed total, the larger rate can make the worst case occur at the smaller reading.
- Hint 1
-
Problem 9 Two explanations for the tank reading
A model tank and its full-size version have similar interiors and corresponding fill marks at a model-to-actual length scale of 1:5. The model holds 104 cubic centimeters to the lower mark and 176 cubic centimeters to the upper mark.
A display intended to show the mass of liquid alone reads 12.61 kg at the lower mark. A technician considers two explanations. In both, the liquid is uniform.
Explanation D. The display is accurate. The liquid's density must be inferred from the lower reading.
Explanation O. The liquid has density 0.92 kilograms per liter. The display adds the same fixed mass offset at any fill level.
For each explanation, find its unknown density or offset and predict the actual liquid mass at the upper mark. In a separate test, the technician fills to the upper mark, drains all the liquid without loss, and weighs it on a calibrated scale. Its exact mass is 20.24 kg. Which explanation is consistent with that test?
The labels give model volumes from the inside floor to each mark. Text description of this figure
A schematic side view shows the upper fill mark above the lower fill mark. In the model, the lower mark is reached with 104 cubic centimeters of liquid and the upper mark with 176 cubic centimeters.
- Hint 1
Find the actual liquid volumes at both fill marks before interpreting the mass reading.
- Hint 2
For D, divide the lower mass by its volume. For O, compare the lower reading with the mass predicted by the stated density.
- Hint 3
The separate scale measures actual mass. The offset applies only to the tank's display.
Answer
Explanation D gives density 0.97 kg/L and predicts an upper-mark mass of 21.34 kg. Explanation O gives a positive offset of 0.65 kg and predicts an upper-mark mass of 20.24 kg. Only Explanation O is consistent with the separate weighing.
Full solution
Actual volume is 125 times model volume.
The lower volume in cubic centimeters is
The lower mark holds 13 L.
The upper volume in cubic centimeters is
The upper mark holds 22 L.
Explanation D.
The density in kilograms per liter is
The actual liquid mass at the upper mark in kilograms is
Explanation O.
The actual liquid mass at the lower mark in kilograms is
The display offset in kilograms is
Explanation O.
The actual liquid mass at the upper mark in kilograms is
The offset belongs to the tank display, so it does not enter the actual mass prediction.
The lower reading can be explained either way.
The separate weighing at the upper mark matches only Explanation O.
Answer
Explanation D gives density 0.97 kg/L and predicts an upper-mark mass of 21.34 kg. Explanation O gives a positive offset of 0.65 kg and predicts an upper-mark mass of 20.24 kg. Only Explanation O is consistent with the separate weighing.
Key idea
A second measurement can distinguish models that were both fitted to the first reading.
- Hint 1
-
Problem 10 A pair of pens along a wall
A long straight wall forms the back of two side-by-side rectangular pens of equal depth. Use new fencing across the full front, along the two outside ends, and along the divider between the pens. Each end and the divider meets the wall at a right angle. There are 66 m of fencing available, with no waste or gaps. Any positive real dimensions are allowed within the stated limits. Both pens must have positive widths, and the wall is long enough for every feasible design.
Original site. The pens may be at most 8 m deep. Find the overall span along the wall and the depth that give the largest combined area. Prove that no feasible design has greater area.
Expanded site. The depth limit rises to 13 m. Find and justify the new optimum. How much combined area does the change gain?
The dashed divider needs new fencing, as do the front and the two ends. Text description of this figure
Two rectangles share a divider from the wall to the front fence. The wall forms their backs. The two outside ends and the divider each span the common depth d. The entire front fence has length w. No pen widths are specified.
- Hint 1
For a fixed depth, a longer front increases the total area. How long can the front be when the two ends and divider have all been paid for in fencing?
- Hint 2
After expressing the span in terms of depth, write the resulting area as a constant minus a positive multiple of a square.
- Hint 3
Compare the depth closest to the square's center under each site limit. Then check that both pens can have positive widths at the chosen span.
Answer
Original site. The optimum has span 42 m and depth 8 m, with combined area 336 square meters. Expanded site. The optimum has span 33 m and depth 11 m, with combined area 363 square meters. The gain is 27 square meters.
Full solution
Let be the span along the wall and be the depth, in meters.
The front uses meters.
The two ends and divider use meters, so
For any feasible depth, area increases with .
The best span at that depth uses the full allowance.
This is positive throughout both sites' allowed depth ranges.
At that span the combined area is
Completing the square gives
Original site.
For , the distance from to 11 is at least 3.
Thus
The largest area is at most 336 square meters.
At the span is 42 m, and a divider at the midpoint gives two positive widths.
This design attains the bound.
Expanded site.
The term is nonnegative for every depth.
Now is allowed, so this term can be zero.
The span is 33 m, and a midpoint divider again gives two positive widths.
This design attains the upper bound of 363 square meters.
The gain in square meters is
Answer
Original site. The optimum has span 42 m and depth 8 m, with combined area 336 square meters. Expanded site. The optimum has span 33 m and depth 11 m, with combined area 363 square meters. The gain is 27 square meters.
Key idea
A bound on depth can put the best design at a site limit until that limit moves past the unconstrained optimum.
- Hint 1