Big Ideas Math: Modeling Real Life, Grade 8
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1. Volumes of Cylinders
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Exercise 27 Page 432

Practice makes perfect

We want to make cylindrical candles of 3 different sizes to sell. Let's call them A, B, and C. To make it a bit easier, we can make them all with the same radius and only vary their heights. Let's also make the heights of the larger candles 2 and 3 times the height of the smallest.

We need to make 8 candles of each size and we have a total of 1 cubic foot of candle wax to use to create them. This gives us an equation for the total volume. \begin{gathered} 8V_\text{A}+8V_\text{B}+8V_\text{C} = 1 \text{ ft}^3 \end{gathered}Let's convert the total volume to cubic inches using a conversion factor. 1 ft^3* 1728 in^3/1 ft^3=1728 in^3 We can substitute the formula for the volume of a cylinder into our equation and solve for h. Remember, the height for the cylinders are 3h, 2h, and h.
8V_\text{A}+8V_\text{B}+8V_\text{C} = 1728
8(\pi {\color{#0000FF}{r}}^2 {\color{#009600}{3h}})+8V_\text{B}+8V_\text{C} = 1728
8(\pi r^2 3h)+8(\pi {\color{#0000FF}{r}}^2 {\color{#009600}{2h}})+8V_\text{C} = 1728
8(π r^2 3h)+8(π r^2 2h)+8(π r^2 h) = 1728
Solve for h
8π r^2 3h+8π r^2 2h+8π r^2 h = 1728
8π r^2(3h+2h+h)= 1728
3h+2h+h=1728/8π r^2
6h=1728/8π r^2
h=1728/8π r^2*6
h=1728/48π r^2
h=48*36/48π r^2
h=48*36/48π r^2
h=36/π r^2
Now we can choose a value for the radius and substitute it. Let's try 2.5 inches.
h=36/π r^2
h=36/π ( 2.5)^2
h=36/π (6.25)
h=1.833464...
h≈1.8 in
The height of Candle C should be 1.8 inches. Remember that we cannot round up since it will make us go over the total amount of wax we have. Rounding down means we will not use all of the wax, which is fine. Using the height of Candle C, we can find the height of Candle A and B as well.
Candle Radius (inches) Height (inches) Volume (cubic inches)
A 2.5 3*1.8= 5.4 π ( 2.5)^2 ( 5.4) ≈ 106
B 2.5 2*1.8= 3.6 π ( 2.5)^2 ( 3.6) ≈ 70.7
C 2.5 1.8 π ( 2.5)^2 ( 1.8) ≈ 35.3

Please remember that these are only example dimensions for the candles. There are many possible combinations of candles sizes that would fit the given requirements.

Now we want to decide on a price for each candle to make a profit of $100. The profit is equal to sales revenue minus the cost of the goods sold. We know the cost of the goods is $20, so we can calculate the total amount of sales revenue we need.
profit=sales-costs
$ 100=sales- $ 20
$ 120=sales
sales= $120
To make a $100 profit, we need to sell the candles for a total of $120. Since we decided on a relation between the heights of the candles, and they all have the same radius, the same relation is true for the volumes. \begin{gathered} V_\text{A}=\pi {\color{#0000FF}{r}}^2 {\color{#009600}{3h}} \quad \Leftrightarrow\quad 3(\pi {\color{#0000FF}{r}}^2 {\color{#009600}{h}})=3V_\text{C}\\ V_\text{B}=\pi {\color{#0000FF}{r}}^2 {\color{#009600}{2h}} \quad \Leftrightarrow\quad 2(\pi {\color{#0000FF}{r}}^2 {\color{#009600}{h}})=2V_\text{C} \end{gathered} We can use this to split the price of each candle based on their volume. Let's call the price of Candle C p. Then the price should be 2p for Candle B and 3p for Candle A. This gives us an equation for finding p.
8(3p)+8(2p)+8(p) = $120
24p+16p+8p=$120
48p=$120
p=$2.50
The price for Candle C should be $2.50. We can calculate the price for Candle A and B as well.
Candle Price
A 3*$2.50=$7.50
B 2*$2.50=$5
C $2.50

Just as with the sizes of the candles, there are many combinations of prices that would fit the given requirements. This is just one example of prices.