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# DVD Profits, Variation 1

## Task

- DVDs can be made in a factory in New Mexico at the rate of 20 DVDs per \$3, but the factory costs \$80,000 to build. If they make 1 million DVDs, what is the unit cost per DVD?
- DVDs can be made in a factory in Colorado at the rate of 10 DVDs per \$1.50, but the factory costs \$100,000 to build. If they make 1 million DVDs, what is the unit cost per DVD?
- How much can a buyer save on a million DVDs by buying DVDs from New Mexico instead of DVDs from Colorado?
- Find an equation for the cost of making $x$ number of DVDs in the factory in New Mexico.
- Find an equation for the cost of making $x$ number of DVDs in the factory in Colorado.

## IM Commentary

The first two problems ask for the unit cost per DVD for making a million DVDs. Even though each additional DVD comes at a fixed price, the overall cost per DVD changes with the number of DVDs produced because of the startup cost of building the factory.

## Solutions

Solution: Working with unit rates

If DVDs didn't need a factory, the DVDs from New Mexico would
just be 20 DVDs per \$3. The cost per DVD would be \$.15
per DVD, or 15 cents per DVD. However, we have to factor
in the factory cost.

The DVDs from Colorado, without the factory, would be 10 DVDs per
\$1.50, so the unit cost would also be \$.15 per DVD, or 15 cents
per DVD, the same as in the factory in New Mexico!
Consequently, the only difference in the cost of
DVDs between New Mexico and Colorado is the cost of the factory,
which costs \$20,000 more in Colorado, no matter how many DVDs are
involved.

- The cost of 1 million DVDs from New Mexico is $$80,000 \text{ dollars } + 1,000,000 \text{ DVDs } \cdot \frac{3 \text{ dollars }}{20 \text{ DVDs }}$$ $$ = 80,000 \text{ dollars } + 50,000 \cdot 3 \text{ dollars }$$ $$ = 80,000 \text{ dollars } + 150,000 \text{ dollars } = 230,000 \text{ dollars }$$ The unit cost per DVD is: $$ \frac{230,000 \text{ dollars } }{1,000,000 \text{ DVDs }} = .23 \frac{\text{ dollars } }{\text{ DVD }} = .23 \frac{\text { dollars }}{\text{ DVD }} \cdot \frac{100 \text{ cents }}{\text{ dollar }} = 23 \frac{\text{ cents }}{\text{ DVD }}$$ So the unit cost is 23 cents per DVD.
- The cost of 1 million DVDs from Colorado is $$ 100,000 \text{ dollars } + 1,000,000 \text{ DVDs } \cdot \frac{1.5 \text{ dollars }}{10 \text{ DVDs }}$$ $$= 100,000 \text{ dollars } + 100,000 \cdot 1.5 \text{ dollars }$$ $$ = 100,000 \text{ dollars } + 150,000 \text{ dollars } = 250,000 \text{ dollars } $$ The unit cost per DVD is: $$ \frac{250,000 \text{ dollars } }{1,000,000 \text{ DVDs }} = .25 \frac{\text{ dollars } }{\text{ DVD }} = .25 \frac{\text { dollars }}{\text{ DVD }} \cdot \frac{100 \text{ cents }}{\text{ dollar }} = 25 \frac{\text{ cents }}{\text{ DVD }} $$ So the unit cost is 25 cents per DVD.
- The difference is 2 cents per DVD. For a million DVDs, the savings would be: $$1,000,000 \text{ DVDs } \cdot\frac{2 \text{ cents }}{\text{ DVD }} \cdot \frac{ \text{ dollar }}{100 \text{ cents }} = 20,000 \text{ dollars }.$$ Of course, the rate of 10 DVDs per 1.5 dollars is the same as 20 DVDs per 3 dollars, so the only difference is the startup cost difference, \$20,000, no matter how many DVDs are made, 0 on up! We have also already computed the cost of a million DVDs from each state, so all we really had to do was subtract the cost of the DVDs from Colorado from the cost of the DVDs from New Mexico, $$ 250,000 \text{ dollars } - 230,000 \text{ dollars } = 20,000 \text{ dollars }. $$
- The cost $C$ of making $x$ DVDs in New Mexico is $$C = 80,000 \text{ dollars } + \frac{ 3 \text{ dollars }}{20 \text{ DVDs }} \cdot x \text{ DVDs }$$ $$C = (80,000 + \frac{3}{20} x) \quad \text{ dollars }$$ $$C = (80,000 + .15 x) \quad \text{ dollars}.$$
- The cost $C$ for making $x$ DVDs in Colorado is $$C = 100,000 \text{ dollars } + \frac{ 1.5 \text{ dollars }}{10 \text{ DVDs }} \cdot x \text{ DVDs } $$ $$C =(100,000 + \frac{1.5}{10} x) \quad \text{ dollars } $$ $$C =(100,000 + .15 x) \quad \text{ dollars}. $$

Solution: Working backwards

If we find the equations asked for in parts (d) and (e) first, we might have saved ourselves a little
work for parts (a) and (b).

If we make $x$ DVDs and we want the unit cost per DVD, we could just use the equation for the cost for $x$ DVDs in New Mexico:
$$\frac{(80,000 + .15 x ) \quad \text{ dollars }}{ x \text{ DVDs }}=
\left(\frac{80,000}{x} + .15\right) \frac{\text{ dollars }}{\text{ DVD }}.
$$
Or, in terms of cents:
$$\left(\frac{80,000}{x} + .15\right) \frac{\text{ dollars }}{\text{ DVD }}
\times \frac{100 \text{ cents }}{\text{ dollar }}=
\left(\frac{8,000,000}{x} + 15 \right) \frac{\text{ cents }}{\text{ DVD }}.
$$
Now, to do part (a), we just plug in 1,000,000
and we get that one DVD costs (8+15) cents = 23 cents.

If we make $x$ DVDs and we
want the unit cost per DVD in Colorado, we could use the equation for the cost for $x$ DVDs in Colorado:
$$\frac{(100,000 + .15 x) \quad \text{ dollars }}{ x \text{ DVDs }}=
\left(\frac{100,000}{x} + .15\right) \frac{\text{ dollars }}{\text{ DVD }}.
$$
Or, in terms of cents:
$$\left(\frac{100,000}{x} + .15\right) \frac{\text{ dollars }}{\text{ DVD }}
\times \frac{100 \text{ cents }}{\text{ dollar }}=
\left(\frac{10,000,000}{x} + 15 \right) \frac{\text{ cents }}{\text{ DVD }}.
$$
Now, to do part (b), we just plug in 1,000,000
and we get that one DVD costs (10+15) cents = 25 cents.

## DVD Profits, Variation 1

- DVDs can be made in a factory in New Mexico at the rate of 20 DVDs per \$3, but the factory costs \$80,000 to build. If they make 1 million DVDs, what is the unit cost per DVD?
- DVDs can be made in a factory in Colorado at the rate of 10 DVDs per \$1.50, but the factory costs \$100,000 to build. If they make 1 million DVDs, what is the unit cost per DVD?
- How much can a buyer save on a million DVDs by buying DVDs from New Mexico instead of DVDs from Colorado?
- Find an equation for the cost of making $x$ number of DVDs in the factory in New Mexico.
- Find an equation for the cost of making $x$ number of DVDs in the factory in Colorado.

## Comments

Log in to comment## Ashley Williams says:

almost 5 yearsI'm having difficulty with the alignment provided. Since there is no graphing in this task, perhaps it belongs in 7th grade?

## Kristin says:

almost 5 yearsGood question. This task is aligned with the cluster "Understand the connections between proportional relationships, lines, and linear equations" and focuses on the connection between proportional relationships and linear equations, but not on the connection between graphs and these other two things. Not all tasks that students do need to draw on or address all aspects of a standard or cluster, but the set of tasks students work on taken as a whole needs to get at all the different angles, so this task isn't enough to illustrate the cluster all by itself.

## Ashley Williams says:

almost 5 yearsGot it! Thanks for clarifying!

## Kristin says:

almost 5 yearsThanks for your comment--I bet a lot of people found this puzzling!