7. Modeling in Three Dimensions
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Chapter 3
7.
Modeling in Three Dimensions
This lesson covers the essentials of three-dimensional modeling, focusing on its practical applications. It teaches how to use geometric shapes like cylinders, cones, and spheres for calculating volumes and surface areas. These skills prove useful in various scenarios, such as determining the capacity of a juice glass at a fair or the water volume an aquarium can hold. The lesson even explores complex calculations like estimating the number of grains of sand a hand can hold. Overall, the aim is to equip individuals with the mathematical tools needed for making informed decisions in everyday situations.
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Lesson Settings & Tools
| | 8 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Modeling in Three Dimensions
Slide of 8
Many objects used in daily life can be modeled as three-dimensional figures. In this lesson, geometric figures will be used to describe some daily life objects.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Comparing Aquarium Volumes
Dominika wants to buy a new aquarium for her fish. She is interested in two types of aquariums that have equal linear measurements.
a What is the volume of Aquarium A? If necessary, round the answer to the nearest cubic foot.
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b For each aquarium, what is the area of the water’s surface when filled to a height of h feet? Write the answer in terms of h and π.
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c Use the answers found above to find the volume of the other aquarium.
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Example
Comparing Aquarium Volumes II
Recall how the formula for the volume of a sphere is proven. The same thought process used in the proof can be applied to solve the challenge.
Two types of aquariums attract the attention of Dominica: Aquarium A and Aquarium B.
a What is the volume of Aquarium A? Round the answer to the nearest cubic foot.
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b For each aquarium, what is the area of the water’s surface when filled to a height of h feet? Write the answer in terms of h and π.
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c Use the answers found above and find the volume of the other aquarium.
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Hint
a Subtract the volume of the cone from the volume of the cylinder.
b Examine how the cross-sections that are parallel to the bases change.
c What does the Cavalieri's Principle states?
Solution
a By subtracting the volume of the cone from the volume of the cylinder, the volume of the Aquarium A can be found. Aquarium A && Cylinder && Cone [0.8em]
V_A& =& V_1& - &V_2
Recall the formulas for the volumes of a cylinder and a cone.
Aquarium A can hold about 262 cubic feet of water.
| Volume of the Cylinder | Volume of the Cone | |
|---|---|---|
| Formula | V_1 = π r^2 h | V_2 = 1/3π r^2 h |
Both the cone and the cylinder forming Aquarium A have a 10 foot diameter, and therefore both have a radius of 5 feet. With that in mind, substitute r= 5 and h= 5.
| Volume of the Cylinder | Volume of the Cone | |
|---|---|---|
| Formula | V_1 = π r^2 h | V_2 = 1/3π r^2 h |
| Substitute Values | V_1 = π ( 5)^2 5 | V_2 = 1/3π ( 5)^2 5 |
| Calculate | V_1 = 125 π | V_2 = 125 π/3 |
The difference between V_1 and V_2 will give the number of cubic feet of water that Aquarium A can hold.
V_A=V_1 - V_2
SubstituteII
V_1= 125 π, V_2= 125π/3
V_A= 125 π - 125π/3
▼
Evaluate right-hand side
NumberToFrac
a = 3* a/3
V_A = 375π/3 - 125 π/3
SubFrac
Subtract fractions
V_A = 250π/3
UseCalc
Use a calculator
V_A = 261.799387 ...
RoundInt
Round to nearest integer
V_A ≈ 262
b Begin by examining the cross-sections of each aquarium. Then, write an expression for the area of the water’s surface at height h. 
As seen on the diagram, the vertical cross-sections of the water — the shaded triangles — are isosceles right triangles, and the horizontal cross-sections form two concentric circles. Using the horizontal cross-sections, the area of the water’s surface at height h will be found. To do so, subtract the area of the inner circle A_(inner) from the area of the outer circle A_(outer).
When Aquarium A is filled to a height of h feet, the area of the water’s surface is 10 π h - π h^2 square feet.
The diagram shows a right triangle ABC. Its hypotenuse and length of one of its legs is known in terms of h. Therefore, using the Pythagorean Theorem, BC can also be expressed in terms of h.
Now, considering the horizontal cross-sections of Aquarium B, BC will be the radius of the inner circle. 
In this instance, the area of the water’s surface at height h is the area of a circle with a radius of sqrt(10h–h^2).
For Aquarium B, the area of the water’s surface at height h is the same as the other aquarium, 10π h−π h^2 square feet.
Aquarium A
Since the cone inside the aquarium is a right cone, its vertex is directly above the center of its base. Furthermore, its height and radius have the same length. Therefore, an isosceles right triangle inside the cone can be formed, as indicated in the diagram.With this in mind, consider the vertical and horizontal cross-sections of the aquarium.
A_(outer) - A_(inner)
SubstituteValues
Substitute values
π (5)^2 - π (5-h)^2
▼
Evaluate
CalcPow
Calculate power
π (25) - π (5-h)^2
CommutativePropMult
Commutative Property of Multiplication
25 π - π (5-h)^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
25 π - π (25-10h+h^2)
Distr
Distribute - π
25π - 25 π + 10π h - π h^2
SubTerm
Subtract term
10π h - π h^2
Aquarium B
Examine the vertical cross-sections of the hemisphere.
AB^2 + BC^2 = AC^2
SubstituteII
AB= 5-h, AC= 5
( 5-h)^2 + BC^2 = 5^2
▼
Solve for BC
SubEqn
LHS-(5-h)^2=RHS-(5-h)^2
BC^2 = 5^2 - (5-h)^2
CalcPow
Calculate power
BC^2 = 25 - (5-h)^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
BC^2 = 25 - (25-10h+h^2)
Distr
Distribute - 1
BC^2 = 25 - 25+10h-h^2
SubTerm
Subtract term
BC^2 = 10h-h^2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(BC^2) = sqrt(10h-h^2)
SqrtPowToNumber
sqrt(a^2)=a
BC = sqrt(10h-h^2)
c Recall what the Cavalieri Principle states.
According to the Cavalieri Principle, the volumes of water must be equal when both aquariums are filled. Therefore, no further calculations are necessary — Aquarium B will hold about 262 cubic feet of water.
|
Cavalieri Principle |
|
Two solids with the same height and the same cross-sectional area at every altitude have the same volume. |
Both aquariums have a height of 5 feet, and the area of the water’s surface when filled to a height of h feet is the same for each aquarium.
| Aquarium A | Aquarium B | |
|---|---|---|
| Height (ft) | 5 | |
| Cross-Sectional Area (ft^2) | 10 π h-π h^2 | |
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Example
Finding the Length of a Toilet Paper Roll
Tiffaniqua wants to calculate the length of the a toilet paper roll. Hey! It is on a great sale, Okay. She draws a diagram and denotes the thickness of the paper, the inner radius, and the outer radius by t, r, and R, respectively.
a Write an expression for the length L of the paper roll in terms of t, r, and R.
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b Find the length of the paper roll if t=0.03, r=2.5, and R = 5.5, all measured in centimeters.
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Solution
a Consider the following horizontal cross-section of the toilet roll. The cross-section consists of concentric circles.
The area of the front face is the product of t and L. That product is also equal to A. Therefore, t * L can be substituted for A in the derived formula for the area of the shaded region.
Note that there might be several alternative ways to find the equation. Only one possible way was shown here.
The area of the shaded region A can be calculated in two ways. It can be expressed as the area of the circle of radius R minus the area of the circle of radius r. A = π R^2 -π r^2
Alternatively, it can be expressed as the area of the front face of the long thin rectangular prism, which is created when the paper is unrolled. b By substituting t= 0.03, r= 2.5, and R= 5.5 into the previously derived equation, the length of the paper towel will be calculated.
The length of the paper is about 2513 centimeters.
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Example
Cylindrical Soda Can
A cylindrical soda can is made of aluminum. It is 6 inches high and its bases have a radius of approximately 1.2 inches.
Give a go at answering the following set of questions. If necessary, round the answer to two decimal places.
a Find the surface area of the soda can.
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b The density of aluminum is approximately 2.7 grams per cubic centimeter. If the mass of the soda can is approximately 21 grams, how many cubic centimeters of aluminum does it contain?
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c Suppose that the thickness of the soda can is uniform throughout its body. Estimate the soda can's thickness.
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Hint
a Use the formula for the surface area of a cylinder.
b The density of a substance is equal to its mass divided by its volume.
c How might the surface area and volume of the soda can be related?
Solution
a Since the soda can is a cylinder, the formula for the surface area of a cylinder will be used.
S=2π rh+2π r^2 Here, r and h are the radius and the height of the cylinder, respectively. The radius is 1.2 inches and the height is 6 inches. Substitute these values in the formula and evaluate it.
Therefore, the surface area of the soda can is about 54.29 square inches. b Recall that the density of a substance is equal to its mass divided by its volume. In other words, the volume of a substance is its mass divided by its density.
d = m/V ⇔ V = m/d Since the mass m of the aluminum can and the density d of aluminum are given, the volume can be calculated. Substitute m= 21 and d= 2.7 into the equation.
The soda can is made of about 7.78 cubic centimeters of aluminum. c To determine the thickness of the soda can, its surface area and volume are good to use. Surface Area & & Volume 54.29 in^2 & & 7.78 cm^3
Now, by substituting V = 7.78 and S= 352.885, t can be found.
This means that the thickness of the soda can is about 0.02 centimeters.
Let t denote the thickness. Then, the volume of the aluminum part of the soda can will be equal to the surface area of the soda can multiplied by its thickness. Therefore, to find t, the the aluminum part's volume should be divided by the soda can's surface area. V = S * t ⇔ t = V/S However, since the amount of aluminum is found in cubic centimeters and the surface area of the soda can in square inches, a conversion factor must be used. Use 6.5 cm^2in^2 to convert square inches to square centimeters.
54.29 in^2 * 6.5 cm^2/in^2
▼
Simplify
MoveLeftFacToNum
a*b/c= a* b/c
54.29 in^2 * 6.5 cm^2/in^2
CrossCommonFac
Cross out common factors
54.29 in^2 * 6.5 cm^2/in^2
SimpQuot
Simplify quotient
54.29 * 6.5 cm^2
Multiply
Multiply
352.885 cm^2
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Example
Choosing Type of Glass
By modeling real-life objects using geometric shapes, various characteristics of the objects can be determined. These characteristics can then be compared to make inferences which could impact real decisions to be made.
Emily is attending a fair and wants to sell 1.5 liters of homemade orange juice she is naming Oranjya Thirsty. She needs to decide the type of glass she will use to serve the juice — a cocktail glass or a Collins glass.
A cocktail glass is a type of glass that has an inverted cone bowl. The cone bowl's height is 5.8 centimeters and the radius of its base is 5.4 centimeters. A collins glass is a cylindrical glass with a height of 9 centimeters and a radius of 3.2 centimeters. Help Emily make a decision by answering the following questions.
a How many cocktail glasses of orange juice can she sell?
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b How many Collins glasses of orange juice can she sell?
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c If Emily chooses cocktail glasses, she will sell each for 6 dollars. If the Collins glasses are chosen, each will be sold for 10 dollars. Which glass type selection makes Emily more money?
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Hint
a Use the formula for the volume of a cone to determine how many liters of orange juice a cocktail glass can hold.
b Use the formula for the volume of a cylinder to determine how many liters of orange juice a Collins glass can hold.
c Use the answers found in the previous parts.
Solution
a First, the volume of the bowl of a cocktail glass will be calculated. Then, it will be used to find the number of glasses. The bowl of a cocktail glass can be modeled by a cone as shown.
This means that a cocktail glass can hold approximately 177 cubic centimeters orange juice. Conversion needs to be made. The conversion factor 1 L1000 cm^3 will convert cubic centimeters into liters.
177 cm^3 * 1 L/1000 cm^3 = 0.177 L
Finally, the number of cocktail glasses can be calculated.
Therefore, 1.5 liters of orange juice fully fills 8 cocktail glasses.
The height of the cone bowl and the radius of its base are 5.8 and 5.4 centimeters, respectively. Substitute these values into the volume formula of a cone.
V = 1/3 π r^2 h
SubstituteII
r= 5.4, h= 5.8
V = 1/3 π ( 5.4)^2 5.8
▼
Evaluate right-hand side
CalcPow
Calculate power
V = 1/3 π (29.16) 5.8
Multiply
Multiply
V = 1/3 169.128 π
MoveRightFacToNumOne
1/b* a = a/b
V = 169.128 π/3
UseCalc
Use a calculator
V = 177.110427 ...
RoundInt
Round to nearest integer
V ≈ 177
b Similarly, start by finding the volume of the cylindrical glass.
Its height is 9 centimeters and its base radius is 3.2 centimeters. Substitute these values into the formula.
This means that the volume of a collins glass is about 290 cubic centimeters. After converting its unit of measurement into liters, it is 0.29 liters. Now, the volume of a carton of orange juice can be divided by the volume of a collins glass. 1.5/0.29 = 5.172413 ... Therefore, 1.5 liters of orange juice fully fills 5 collins glasses. c Emily can sell 8 cocktail glasses or 5 collins glasses. Knowing the prices, the revenue from each sale can be calculated.
As a result, Emily should choose the Collins glass, as the revenue from this selection is greater.
| Type of Glass | |
|---|---|
| Cocktail Glass | Collins Glass |
| 8 * 6 | 5 * 10 |
| $48 | $50 |
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Example
Estimating How Many Grains of Sand a Hand Can Hold
With the help of geometric modeling, any number of objects can be approximated regardless of whether they are super large or tiny minuscule grains of sand.
Take, for example, Ramsha's situation. She is looking through photos from her trip to the beach to post on her social media page. A photo that shows her holding sand sparks her curiosity. She wonders how many individual grains of sand is she holding. Ramsha thinks she can model a grain of sand using a sphere. She then assumes that each grain has a diameter of 0.06 centimeters.
Answer
About 1.7 * 10^6
Hint
The formula for the volume of a sphere is V = 43π r^3, where r is the radius of the sphere.
Solution
To find the number N of grains of sand in Ramsha's hands, the mass M of the sand in her hands should be divided by the mass m of a grain.
N = M/m
Recall that the density of a substance is equal to its mass divided by its volume. In other words, the mass of a substance is its density times its volume. d= m/V ⇔ m = d * V
Since the density of a grain of sand is given, the volume of a grain of sand will be calculated first.
Finding the Volume of a Grain
The radius of a grain is 0.062=0.03 centimeters. Use the formula for the volume of a sphere to find the volume of a grain.
V = 4/3 π r^3
Substitute
r= 0.03
V = 4/3 π ( 0.03)^3
▼
Evaluate right-hand side
CalcPow
Calculate power
V = 4/3 π (0.000027)
CommutativePropMult
Commutative Property of Multiplication
V = 4/3 0.000027 π
MoveRightFacToNum
a/c* b = a* b/c
V = 0.000108 π/3
UseCalc
Use a calculator
V = 0.000113 ...
RoundDec
Round to 5 decimal place(s)
V ≈ 0.00011
Write in scientific notation
V ≈ 1.1 * 10^(- 4)
The volume of a grain is about 1.1 * 10^(- 4) cubic centimeters.
Finding the Mass of a Grain
The density of a grain is 1.52 g/cm^3 and its volume is 1.1 * 10^(- 4) cubic centimeters. By multiplying these values, the mass of a grain can be found.
1.52 g/cm^3 * 1.1 * 10^(- 4) cm^3
▼
Evaluate
MoveRightFacToNum
a/c* b = a* b/c
1.52 g * 1.1 * 10^(- 4) cm^3/cm^3
CancelCommonFac
Cancel out common factors
1.52 g * 1.1 * 10^(- 4) cm^3/cm^3
SimpQuot
Simplify quotient
1.52 g * 1.1 * 10^(- 4)
CommutativePropMult
Commutative Property of Multiplication
1.52 * 1.1 * 10^(- 4) g
Multiply
Multiply
1.672 * 10^(- 4) g
RoundDec
Round to 1 decimal place(s)
1.7 * 10^(- 4) g
Finding the Number of Grains
Finally, substitute the values into the formula mentioned at the beginning to calculate the number of grains of sand.
N = M/m
SubstituteII
M= 300 g, m= 1.7 * 10^(- 4) g
N = 300 g/1.7 * 10^(- 4) g
▼
Evaluate right-hand side
CrossCommonFac
Cross out common factors
N = 300 g/1.7 * 10^(- 4) g
SimpQuot
Simplify quotient
N = 300/1.7 * 10^(- 4)
WriteProdFrac
Write as a product of fractions
N = 300/1.7 ( 1/10^(- 4) )
FracToNegExponent
1/a^m=a^(- m)
N = 300/1.7 (10^4)
CalcQuot
Calculate quotient
N = (176.470588 ... )(10^4)
Multiply
Multiply
N = 1 764 705.882352 ...
RoundSigDig
Round to 2 significant digit(s)
N ≈ 1 700 000
Write in scientific notation
N ≈ 1.7 * 10^6
The number of grains of sand is approximately 1.7 * 10^6, or about 1.7 million.
the total number of stars in the universe is greater than all the grains of sand on all the beaches of the planet Earth.
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Closure
Modeling the Human Eye
Research projects usually require an interdisciplinary approach. That is, people from different disciplines work together to develop and test hypothesis, run experiments, and test theories.
Biologists, for example, can work with mathematicians to model a part of an organism. By doing so, researchers can predict how these parts function, grow, and change. For example, the human eye was able to be modeled as a sphere. Move the slider to rotate the eye.
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Modeling in Three Dimensions
Exercise 1.1
A cylinder-shaped cistern has an inner diameter of 10 meters and currently holds 1200 cubic meters of oil. The oil fills up two-thirds of the cistern. What is the difference between the height the oil reaches and the height of the cistern? Round the difference to one decimal place.
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Examining the diagram, we see that the oil is in itself a cylinder with the same inner diameter as the cistern. Since we have been given the volume of the oil, we can determine how high it reaches by using the formula for the volume of a cylinder and solving for the height h. V=π r^2 h
Height of the Oil
Let's substitute the volume, 1200 cubic centimeters and radius, 102=5 meters, into the formula to solve for the height the oil reaches.
We will keep the height of the oil in exact form to maintain precision accuracy in our calculations.
Height of the Cistern
We know that the oil reaches two-thirds of the cylinder's height. If we call the cylinder's height H, we can write the following equation. 2/3H=h Let's substitute the value of h and solve for H.
The height of the cistern is 72π meters.
Difference Between Heights
To determine the height from the top of the oil to the top of the cistern, we can subtract the height where the oil reaches from the height of the cistern.
The difference between the two variables is about 7.6 meters.
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Hint & Answer
S
Solution
A cube of copper has side lengths of 20 centimeters. How many smaller cubes can be made from the original cube if each of the smaller cubes will have a side length of 2 centimeters?
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Each side length is 20 centimeters. Divide that number by 2, which is the target length of the new smaller cubes. This tells us that we can fit ten 2-centimeter segments. Therefore, along each dimension, we would be able to fit 10 small cubes.
If we multiply the number of small cubes that fit along three of the original cube's sides, we get the total number of small cubes that can be created from the original cube. (10)(10)(10)=1000
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Hint & Answer
S
Solution
A silo for wheat consists of a cylinder and a hemisphere.
The total height of the silo is 30 meters, and it has a radius of 5 meters. How many cubic meters of space are available to hold wheat? Round the amount to the nearest cubic meter.
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From the given information, we know that the radius of the cylinder is 5 meters. This means that the cylinder's height is 25 meters.
To calculate the volume of this composite solid, we have to determine the volume of the cylinder and of the hemisphere then add the results.
Volume of the Hemisphere
To calculate the volume of the hemisphere V_(HS), we divide the formula for finding the volume of a sphere by 2. V_(HS) = 43π r^3/2 Let's substitute the hemisphere's radius into the formula and evaluate.
The volume of the hemisphere is 250π3 cubic meters.
Volume of the Cylinder
To calculate the volume of the cylinder V_C, we multiply the base area, which is a circle, by its height.
The cylindrical part of the silo has a volume of 625π cubic meters.
Volume of Silo
Now that the volume of each part is known, we can determine the silo's total volume by adding its parts.
The volume of the silo is about 2225 cubic meters. That is the amount of available space to hold wheat or any other substance.
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S
Solution
Which box gives the most cereal in relation to its cost?
.png)
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To compare the prices, we have to calculate the amount of cubic inches of cereal we get per dollar for each of the boxes. In other words, we want to divide the volume of each box by their respective price. Volume of Cereal/Price We will start by finding the volume of each box.
Volume of the Larger Box
Let's calculate the volume of the large box. This is a rectangular prism. To find its volume, we multiply, its width, length and height. From the diagram, we can see w= 3, l = 8, and h= 10. Let's multiply them.
The volume of the large box is 240 cubic inches.
Volume of the Smaller Box
We will calculate the volume of the small box in the same way. From the diagram we can see that the dimensions are w= 2, l = 6, and h= 8.
The volume of the small box is 96 cubic inches.
Amount of Cereal in Cubic Inches per Dollar
Now we can find how many cubic inches of cereal we get per dollar by dividing the volume of each box by the respective price. Large box:& 240/6=40in.$^3$ per dollar [1em] Small box:& 96/3=32in.$^3$ per dollar With the large box, we get 40 cubic inches of cereal per dollar as opposed to 32 cubic inches of cereal per dollar when buying the smaller box. Therefore, buying the larger box is the better deal in terms of cost and amount of cereal.
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A jeweler is offering to sale Zain either seven golden marbles with a radius of 4 millimeters each, or one golden marble with a radius of 8 millimeters? Which offer should Zain choose if they want the most amount of gold?
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To compare the offers, we will calculate the combined volume of the seven smaller marbles and then the volume of the giant marble. It appears that the offer with the bigger volume of gold is better. Let's check if this is the case. Since these are spheres, we will use the formula for the volume of a sphere. V=4/3π r^3
Total Volume of Seven Smaller Marbles
Before calculating the total volume of the smaller marbles, we will calculate the volume of one of them. Then, we will multiply it by 7 to account for all of the marbles.
Finally, we multiply the volume of one marble by 7 to get the total volume of the seven marbles. 7V=7(256π/3)=1792π/3
Volume of the Giant Marble
Again, we will use the same formula to calculate the volume of the giant marble. To do so, we will substitute r=8 into the formula.
Comparison
Since we have found both volumes, we can compare the offers. Small marbles:& V=1792π/3 [0.75em] Giant marble:& V=2048π/3 Since the volume of the giant marble has a greater numerator, Zain should choose the giant marble to obtain the most amount of gold.
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Modeling in Three Dimensions
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