Math / Calculus

QuestionQuestion The marginal cost of manufacturing $x$ units of a certain product is $C^{\prime}(x)=\frac{x^{2}}{2}-2 x+29$ , in dollars per unit. Find the increase in cost if the production level is raised from 4 units to 8 units.
Enter your answer as a decimal, rounded to the nearest cent.
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Studdy Solution

STEP 1

What is this asking? How much more money does it cost to make 8 gizmos than to make 4 gizmos, if we know how much the cost changes at each step? Watch out! We're given the rate of change of the cost, not the cost itself!

STEP 2

1. Find the cost function
2. Calculate the cost difference

STEP 3

We're given the marginal cost $C^{\prime}(x) = \frac{x^2}{2} - 2x + 29$ , which tells us how much the cost changes as we make more gizmos.
To find the total cost of making $x$ gizmos, we need to find a function $C(x)$ whose derivative is $C^{\prime}(x)$ .
That means we need to integrate the marginal cost function!

STEP 4

Let's integrate $C^{\prime}(x)$ with respect to $x$ : $C(x) = \int C^{\prime}(x) \, dx = \int \left( \frac{x^2}{2} - 2x + 29 \right) dx$ $C(x) = \frac{x^3}{2 \cdot 3} - \frac{2x^2}{2} + 29x + K = \frac{x^3}{6} - x^2 + 29x + K$ where $K$ is the constant of integration.
This constant represents the fixed costs, like renting the factory, which don't change based on how many gizmos we make.

STEP 5

We want to find the increase in cost when production goes from 4 units to 8 units.
This is the difference between the cost of making 8 units and the cost of making 4 units.
We can write this as $C(8) - C(4)$ .

STEP 6

Let's plug in $x = 8$ into our cost function: $C(8) = \frac{\textbf{8}^3}{6} - \textbf{8}^2 + 29 \cdot \textbf{8} + K = \frac{512}{6} - 64 + 232 + K = \frac{256}{3} + 168 + K$

STEP 7

Now let's plug in $x = 4$ : $C(4) = \frac{\textbf{4}^3}{6} - \textbf{4}^2 + 29 \cdot \textbf{4} + K = \frac{64}{6} - 16 + 116 + K = \frac{32}{3} + 100 + K$

STEP 8

Finally, let's find the difference: $C(8) - C(4) = \left( \frac{256}{3} + 168 + K \right) - \left( \frac{32}{3} + 100 + K \right)$ $C(8) - C(4) = \frac{256 - 32}{3} + 168 - 100 + K - K = \frac{224}{3} + 68 = \frac{224 + 204}{3} = \frac{428}{3} \approx 142.67$

STEP 9

The increase in cost is approximately $\$142.67$ .

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Studdy Solution

STEP 1

What is this asking? How much more money does it cost to make 8 gizmos than to make 4 gizmos, if we know how much the cost changes at each step? Watch out! We're given the rate of change of the cost, not the cost itself!

STEP 2

1. Find the cost function
2. Calculate the cost difference

STEP 3

STEP 4

STEP 5

We want to find the increase in cost when production goes from 4 units to 8 units.
This is the difference between the cost of making 8 units and the cost of making 4 units.
We can write this as $C(8) - C(4)$ .

STEP 6

Let's plug in $x = 8$ into our cost function: $C(8) = \frac{\textbf{8}^3}{6} - \textbf{8}^2 + 29 \cdot \textbf{8} + K = \frac{512}{6} - 64 + 232 + K = \frac{256}{3} + 168 + K$

STEP 7

Now let's plug in $x = 4$ : $C(4) = \frac{\textbf{4}^3}{6} - \textbf{4}^2 + 29 \cdot \textbf{4} + K = \frac{64}{6} - 16 + 116 + K = \frac{32}{3} + 100 + K$

STEP 8

Finally, let's find the difference: $C(8) - C(4) = \left( \frac{256}{3} + 168 + K \right) - \left( \frac{32}{3} + 100 + K \right)$ $C(8) - C(4) = \frac{256 - 32}{3} + 168 - 100 + K - K = \frac{224}{3} + 68 = \frac{224 + 204}{3} = \frac{428}{3} \approx 142.67$

STEP 9

The increase in cost is approximately $\$142.67$ .

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Studdy Solution

STEP 1

What is this asking? How much more money does it cost to make 8 gizmos than to make 4 gizmos, if we know how much the cost changes at each step? Watch out! We're given the *rate* of change of the cost, not the cost itself!

STEP 2

1. Find the cost function2. Calculate the cost difference

STEP 3

STEP 4

STEP 5

We want to find the increase in cost when production goes from 4 units to 8 units.This is the *difference* between the cost of making **8 units** and the cost of making **4 units**.We can write this as C(8)−C(4)C(8) - C(4)C(8)−C(4).

STEP 6

Let's **plug in** x=8x = 8x=8 into our cost function: C(8)=836−82+29⋅8+K=5126−64+232+K=2563+168+K C(8) = \frac{\textbf{8}^3}{6} - \textbf{8}^2 + 29 \cdot \textbf{8} + K = \frac{512}{6} - 64 + 232 + K = \frac{256}{3} + 168 + K C(8)=683​−82+29⋅8+K=6512​−64+232+K=3256​+168+K

STEP 7

Now let's **plug in** x=4x = 4x=4: C(4)=436−42+29⋅4+K=646−16+116+K=323+100+K C(4) = \frac{\textbf{4}^3}{6} - \textbf{4}^2 + 29 \cdot \textbf{4} + K = \frac{64}{6} - 16 + 116 + K = \frac{32}{3} + 100 + K C(4)=643​−42+29⋅4+K=664​−16+116+K=332​+100+K

STEP 8

STEP 9

The increase in cost is approximately $142.67\$142.67$142.67.

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What is this asking? How much more money does it cost to make 8 gizmos than to make 4 gizmos, if we know how much the cost changes at each step? Watch out! We're given the rate of change of the cost, not the cost itself!

1. Find the cost function
2. Calculate the cost difference

We want to find the increase in cost when production goes from 4 units to 8 units.
This is the difference between the cost of making 8 units and the cost of making 4 units.
We can write this as $C(8) - C(4)$ .

Let's plug in $x = 8$ into our cost function: $C(8) = \frac{\textbf{8}^3}{6} - \textbf{8}^2 + 29 \cdot \textbf{8} + K = \frac{512}{6} - 64 + 232 + K = \frac{256}{3} + 168 + K$

Now let's plug in $x = 4$ : $C(4) = \frac{\textbf{4}^3}{6} - \textbf{4}^2 + 29 \cdot \textbf{4} + K = \frac{64}{6} - 16 + 116 + K = \frac{32}{3} + 100 + K$

The increase in cost is approximately $\$142.67$ .