Math / Algebra

QuestionLeasing costs, equipment maintenance, salaries, electricity, and marketing cost a car manufacturer an average of $\$ 100000$ per day to produce a certain type of car. Materials, invoices, and shipping cost the manufacturer an average of $\$ 6000$ per car. a. Suppose that the car manufacturer produces and sells $n$ cars of that type per day. Let $C(n)$ be the total daily cost (in dollars). Find a formula of $C$ . $C(n)=100000+6000 n$ b. Suppose that the car manufacturer produces and sells $n$ cars of that type per day. Let $B(n)$ be the amount (in dollars) the manufacturer should charge per car to break even by selling $n$ cars. Find an equation of $B$ . $B(n)=\frac{100000}{n}+6000$ c. Suppose that the car manufacturer produces and sells $n$ cars of that type per day. Let $P(n)$ be the amount (in dollars) the manufacturer should charge per car to make a profit of $\$ 2000$ per car. Find a formula of P. [Hint: Build on your equation from part (b) to find an equation of P.] $P(n)=\frac{100000}{n}+8000$ d. Find $P(40)$ . $\mathrm{P}(40)=\$ \square$ (Round to the nearest dollar.)

Studdy Solution

STEP 1

What is this asking? A car manufacturer has daily costs and per-car costs, and we need to figure out how much they should charge per car to break even and to make a profit. Watch out! Don't mix up the fixed daily costs with the per-car costs!

STEP 2

1. Find the total daily cost.
2. Find the break-even price per car.
3. Find the price per car for a target profit.
4. Calculate the price for a specific production level.

STEP 3

We're given that the daily fixed costs are $\$100000$ , and the per-car cost is $\$6000$ .
If they make $n$ cars per day, the total per-car cost is $6000 \cdot n$ .

STEP 4

So, the total daily cost, $C(n)$ , is the sum of the fixed cost and the variable cost: $C(n) = 100000 + 6000 \cdot n$

STEP 5

To break even, the total amount they make from selling the cars needs to equal the total daily cost.
If they sell $n$ cars at a price $B(n)$ each, they make $B(n) \cdot n$ .

STEP 6

We want this to equal the total daily cost, so we set up the equation: $B(n) \cdot n = C(n)$ $B(n) \cdot n = 100000 + 6000 \cdot n$

STEP 7

To find $B(n)$ , we divide both sides of the equation by $n$ : $B(n) = \frac{100000}{n} + \frac{6000 \cdot n}{n}$ $B(n) = \frac{100000}{n} + 6000$

STEP 8

Now, we want to make a profit of $\$2000$ per car.
This means we need to add $2000 \cdot n$ to our break-even cost.

STEP 9

Let $P(n)$ be the price per car to achieve this profit.
We can build on our break-even price $B(n)$ : $P(n) = B(n) + 2000$ $P(n) = \frac{100000}{n} + 6000 + 2000$ $P(n) = \frac{100000}{n} + 8000$

STEP 10

We need to find $P(40)$ , which means we substitute $n = 40$ into our formula for $P(n)$ : $P(40) = \frac{100000}{40} + 8000$ $P(40) = 2500 + 8000$ $P(40) = 10500$

STEP 11

To make a $\$2000$ profit per car when producing 40 cars per day, the manufacturer should charge $\$10500$ per car.

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Studdy Solution

STEP 1

What is this asking? A car manufacturer has daily costs and per-car costs, and we need to figure out how much they should charge per car to break even and to make a profit. Watch out! Don't mix up the fixed daily costs with the per-car costs!

STEP 2

1. Find the total daily cost.
2. Find the break-even price per car.
3. Find the price per car for a target profit.
4. Calculate the price for a specific production level.

STEP 3

We're given that the daily fixed costs are $\$100000$ , and the per-car cost is $\$6000$ .
If they make $n$ cars per day, the total per-car cost is $6000 \cdot n$ .

STEP 4

So, the total daily cost, $C(n)$ , is the sum of the fixed cost and the variable cost: $C(n) = 100000 + 6000 \cdot n$

STEP 5

To break even, the total amount they make from selling the cars needs to equal the total daily cost.
If they sell $n$ cars at a price $B(n)$ each, they make $B(n) \cdot n$ .

STEP 6

We want this to equal the total daily cost, so we set up the equation: $B(n) \cdot n = C(n)$ $B(n) \cdot n = 100000 + 6000 \cdot n$

STEP 7

To find $B(n)$ , we divide both sides of the equation by $n$ : $B(n) = \frac{100000}{n} + \frac{6000 \cdot n}{n}$ $B(n) = \frac{100000}{n} + 6000$

STEP 8

Now, we want to make a profit of $\$2000$ per car.
This means we need to add $2000 \cdot n$ to our break-even cost.

STEP 9

Let $P(n)$ be the price per car to achieve this profit.
We can build on our break-even price $B(n)$ : $P(n) = B(n) + 2000$ $P(n) = \frac{100000}{n} + 6000 + 2000$ $P(n) = \frac{100000}{n} + 8000$

STEP 10

We need to find $P(40)$ , which means we substitute $n = 40$ into our formula for $P(n)$ : $P(40) = \frac{100000}{40} + 8000$ $P(40) = 2500 + 8000$ $P(40) = 10500$

STEP 11

To make a $\$2000$ profit per car when producing 40 cars per day, the manufacturer should charge $\$10500$ per car.

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Studdy Solution

STEP 1

What is this asking? A car manufacturer has daily costs and per-car costs, and we need to figure out how much they should charge per car to break even and to make a profit. Watch out! Don't mix up the fixed daily costs with the per-car costs!

STEP 2

1. Find the total daily cost.2. Find the break-even price per car.3. Find the price per car for a target profit.4. Calculate the price for a specific production level.

STEP 3

We're given that the daily fixed costs are $100000\$100000$100000, and the per-car cost is $6000\$6000$6000.If they make nnn cars per day, the total per-car cost is 6000⋅n6000 \cdot n6000⋅n.

STEP 4

So, the **total daily cost**, C(n)C(n)C(n), is the sum of the fixed cost and the variable cost: C(n)=100000+6000⋅nC(n) = 100000 + 6000 \cdot nC(n)=100000+6000⋅n

STEP 5

To break even, the total amount they make from selling the cars needs to equal the total daily cost.If they sell nnn cars at a price B(n)B(n)B(n) each, they make B(n)⋅nB(n) \cdot nB(n)⋅n.

STEP 6

We want this to equal the total daily cost, so we set up the equation: B(n)⋅n=C(n)B(n) \cdot n = C(n)B(n)⋅n=C(n) B(n)⋅n=100000+6000⋅nB(n) \cdot n = 100000 + 6000 \cdot nB(n)⋅n=100000+6000⋅n

STEP 7

To find B(n)B(n)B(n), we divide both sides of the equation by nnn: B(n)=100000n+6000⋅nnB(n) = \frac{100000}{n} + \frac{6000 \cdot n}{n}B(n)=n100000​+n6000⋅n​ B(n)=100000n+6000B(n) = \frac{100000}{n} + 6000B(n)=n100000​+6000

STEP 8

Now, we want to make a profit of $2000\$2000$2000 per car.This means we need to add 2000⋅n2000 \cdot n2000⋅n to our break-even cost.

STEP 9

STEP 10

We need to find P(40)P(40)P(40), which means we substitute n=40n = 40n=40 into our formula for P(n)P(n)P(n): P(40)=10000040+8000P(40) = \frac{100000}{40} + 8000P(40)=40100000​+8000 P(40)=2500+8000P(40) = 2500 + 8000P(40)=2500+8000P(40)=10500P(40) = 10500P(40)=10500

STEP 11

To make a $2000\$2000$2000 profit per car when producing 40 cars per day, the manufacturer should charge $10500\$10500$10500 per car.

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1. Find the total daily cost.
2. Find the break-even price per car.
3. Find the price per car for a target profit.
4. Calculate the price for a specific production level.

We're given that the daily fixed costs are $\$100000$ , and the per-car cost is $\$6000$ .
If they make $n$ cars per day, the total per-car cost is $6000 \cdot n$ .

So, the total daily cost, $C(n)$ , is the sum of the fixed cost and the variable cost: $C(n) = 100000 + 6000 \cdot n$

To break even, the total amount they make from selling the cars needs to equal the total daily cost.
If they sell $n$ cars at a price $B(n)$ each, they make $B(n) \cdot n$ .

We want this to equal the total daily cost, so we set up the equation: $B(n) \cdot n = C(n)$ $B(n) \cdot n = 100000 + 6000 \cdot n$

To find $B(n)$ , we divide both sides of the equation by $n$ : $B(n) = \frac{100000}{n} + \frac{6000 \cdot n}{n}$ $B(n) = \frac{100000}{n} + 6000$

Now, we want to make a profit of $\$2000$ per car.
This means we need to add $2000 \cdot n$ to our break-even cost.

We need to find $P(40)$ , which means we substitute $n = 40$ into our formula for $P(n)$ : $P(40) = \frac{100000}{40} + 8000$ $P(40) = 2500 + 8000$ $P(40) = 10500$

To make a $\$2000$ profit per car when producing 40 cars per day, the manufacturer should charge $\$10500$ per car.