Math / Calculus

QuestionFind the cost function C. Determine where the cost is a minimum. (a) $C^{\prime}(x)=14 x-2800$ (b) $C^{\prime}(x)=20 x-8000$
Fixed cost $=\$ 4300$ Fixed cost $=\$ 500$
Provide your answer below:

Studdy Solution

STEP 1

1. We are given two different derivative functions for the cost function $C(x)$ .
2. We need to find the cost function $C(x)$ for each scenario.
3. We need to determine where the cost is minimized for each scenario.
4. The fixed cost is a constant term added to the cost function.

STEP 2

1. Integrate the derivative function $C'(x)$ to find the cost function $C(x)$ .
2. Add the fixed cost to the integrated function.
3. Determine the critical points by setting the derivative $C'(x)$ to zero.
4. Analyze the critical points to find the minimum cost.

STEP 3

For scenario (a), integrate $C'(x) = 14x - 2800$ :
$C(x) = \int (14x - 2800) \, dx$

STEP 4

Perform the integration:
$C(x) = 14 \cdot \frac{x^2}{2} - 2800x + C_0$ $C(x) = 7x^2 - 2800x + C_0$

STEP 5

Add the fixed cost of $4300 to the cost function:
$C(x) = 7x^2 - 2800x + 4300$

STEP 6

Find the critical points by setting $C'(x) = 0$ :
$14x - 2800 = 0$ $14x = 2800$ $x = 200$

STEP 7

Analyze the critical point $x = 200$ to determine if it is a minimum.
Since $C'(x)$ changes from negative to positive at $x = 200$ , it is a minimum.

STEP 8

For scenario (b), integrate $C'(x) = 20x - 8000$ :
$C(x) = \int (20x - 8000) \, dx$

STEP 9

Perform the integration:
$C(x) = 20 \cdot \frac{x^2}{2} - 8000x + C_0$ $C(x) = 10x^2 - 8000x + C_0$

STEP 10

Add the fixed cost of $500 to the cost function:
$C(x) = 10x^2 - 8000x + 500$

STEP 11

Find the critical points by setting $C'(x) = 0$ :
$20x - 8000 = 0$ $20x = 8000$ $x = 400$

STEP 12

Analyze the critical point $x = 400$ to determine if it is a minimum.
Since $C'(x)$ changes from negative to positive at $x = 400$ , it is a minimum.
The cost functions and their minimum points are:
(a) $C(x) = 7x^2 - 2800x + 4300$ with minimum cost at $x = 200$ .
(b) $C(x) = 10x^2 - 8000x + 500$ with minimum cost at $x = 400$ .

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QuestionFind the cost function C. Determine where the cost is a minimum. (a) $C^{\prime}(x)=14 x-2800$ (b) $C^{\prime}(x)=20 x-8000$
Fixed cost $=\$ 4300$ Fixed cost $=\$ 500$
Provide your answer below:

Studdy Solution

STEP 1

1. We are given two different derivative functions for the cost function $C(x)$ .
2. We need to find the cost function $C(x)$ for each scenario.
3. We need to determine where the cost is minimized for each scenario.
4. The fixed cost is a constant term added to the cost function.

STEP 2

1. Integrate the derivative function $C'(x)$ to find the cost function $C(x)$ .
2. Add the fixed cost to the integrated function.
3. Determine the critical points by setting the derivative $C'(x)$ to zero.
4. Analyze the critical points to find the minimum cost.

STEP 3

For scenario (a), integrate $C'(x) = 14x - 2800$ :
$C(x) = \int (14x - 2800) \, dx$

STEP 4

Perform the integration:
$C(x) = 14 \cdot \frac{x^2}{2} - 2800x + C_0$ $C(x) = 7x^2 - 2800x + C_0$

STEP 5

Add the fixed cost of $4300 to the cost function:
$C(x) = 7x^2 - 2800x + 4300$

STEP 6

Find the critical points by setting $C'(x) = 0$ :
$14x - 2800 = 0$ $14x = 2800$ $x = 200$

STEP 7

Analyze the critical point $x = 200$ to determine if it is a minimum.
Since $C'(x)$ changes from negative to positive at $x = 200$ , it is a minimum.

STEP 8

For scenario (b), integrate $C'(x) = 20x - 8000$ :
$C(x) = \int (20x - 8000) \, dx$

STEP 9

Perform the integration:
$C(x) = 20 \cdot \frac{x^2}{2} - 8000x + C_0$ $C(x) = 10x^2 - 8000x + C_0$

STEP 10

Add the fixed cost of $500 to the cost function:
$C(x) = 10x^2 - 8000x + 500$

STEP 11

Find the critical points by setting $C'(x) = 0$ :
$20x - 8000 = 0$ $20x = 8000$ $x = 400$

STEP 12

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QuestionFind the cost function C. Determine where the cost is a minimum. (a) C′(x)=14x−2800C^{\prime}(x)=14 x-2800C′(x)=14x−2800 (b) C′(x)=20x−8000C^{\prime}(x)=20 x-8000C′(x)=20x−8000Fixed cost =$4300=\$ 4300=$4300 Fixed cost =$500=\$ 500=$500Provide your answer below:

Studdy Solution

STEP 1

1. We are given two different derivative functions for the cost function C(x) C(x) C(x).2. We need to find the cost function C(x) C(x) C(x) for each scenario.3. We need to determine where the cost is minimized for each scenario.4. The fixed cost is a constant term added to the cost function.

STEP 2

1. Integrate the derivative function C′(x) C'(x) C′(x) to find the cost function C(x) C(x) C(x).2. Add the fixed cost to the integrated function.3. Determine the critical points by setting the derivative C′(x) C'(x) C′(x) to zero.4. Analyze the critical points to find the minimum cost.

STEP 3

For scenario (a), integrate C′(x)=14x−2800 C'(x) = 14x - 2800 C′(x)=14x−2800:C(x)=∫(14x−2800) dx C(x) = \int (14x - 2800) \, dx C(x)=∫(14x−2800)dx

STEP 4

Perform the integration:C(x)=14⋅x22−2800x+C0 C(x) = 14 \cdot \frac{x^2}{2} - 2800x + C_0 C(x)=14⋅2x2​−2800x+C0​ C(x)=7x2−2800x+C0 C(x) = 7x^2 - 2800x + C_0 C(x)=7x2−2800x+C0​

STEP 5

Add the fixed cost of $4300 to the cost function:C(x)=7x2−2800x+4300 C(x) = 7x^2 - 2800x + 4300 C(x)=7x2−2800x+4300

STEP 6

Find the critical points by setting C′(x)=0 C'(x) = 0 C′(x)=0:14x−2800=0 14x - 2800 = 0 14x−2800=0 14x=2800 14x = 2800 14x=2800 x=200 x = 200 x=200

STEP 7

Analyze the critical point x=200 x = 200 x=200 to determine if it is a minimum.Since C′(x) C'(x) C′(x) changes from negative to positive at x=200 x = 200 x=200, it is a minimum.

STEP 8

For scenario (b), integrate C′(x)=20x−8000 C'(x) = 20x - 8000 C′(x)=20x−8000:C(x)=∫(20x−8000) dx C(x) = \int (20x - 8000) \, dx C(x)=∫(20x−8000)dx

STEP 9

Perform the integration:C(x)=20⋅x22−8000x+C0 C(x) = 20 \cdot \frac{x^2}{2} - 8000x + C_0 C(x)=20⋅2x2​−8000x+C0​ C(x)=10x2−8000x+C0 C(x) = 10x^2 - 8000x + C_0 C(x)=10x2−8000x+C0​

STEP 10

Add the fixed cost of $500 to the cost function:C(x)=10x2−8000x+500 C(x) = 10x^2 - 8000x + 500 C(x)=10x2−8000x+500

STEP 11

Find the critical points by setting C′(x)=0 C'(x) = 0 C′(x)=0:20x−8000=0 20x - 8000 = 0 20x−8000=0 20x=8000 20x = 8000 20x=8000 x=400 x = 400 x=400

STEP 12

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QuestionFind the cost function C. Determine where the cost is a minimum. (a) $C^{\prime}(x)=14 x-2800$ (b) $C^{\prime}(x)=20 x-8000$
Fixed cost $=\$ 4300$ Fixed cost $=\$ 500$
Provide your answer below:

1. We are given two different derivative functions for the cost function $C(x)$ .
2. We need to find the cost function $C(x)$ for each scenario.
3. We need to determine where the cost is minimized for each scenario.
4. The fixed cost is a constant term added to the cost function.

1. Integrate the derivative function $C'(x)$ to find the cost function $C(x)$ .
2. Add the fixed cost to the integrated function.
3. Determine the critical points by setting the derivative $C'(x)$ to zero.
4. Analyze the critical points to find the minimum cost.

For scenario (a), integrate $C'(x) = 14x - 2800$ :
$C(x) = \int (14x - 2800) \, dx$

Perform the integration:
$C(x) = 14 \cdot \frac{x^2}{2} - 2800x + C_0$ $C(x) = 7x^2 - 2800x + C_0$

Add the fixed cost of $4300 to the cost function:
$C(x) = 7x^2 - 2800x + 4300$

Find the critical points by setting $C'(x) = 0$ :
$14x - 2800 = 0$ $14x = 2800$ $x = 200$

Analyze the critical point $x = 200$ to determine if it is a minimum.
Since $C'(x)$ changes from negative to positive at $x = 200$ , it is a minimum.

For scenario (b), integrate $C'(x) = 20x - 8000$ :
$C(x) = \int (20x - 8000) \, dx$

Perform the integration:
$C(x) = 20 \cdot \frac{x^2}{2} - 8000x + C_0$ $C(x) = 10x^2 - 8000x + C_0$

Add the fixed cost of $500 to the cost function:
$C(x) = 10x^2 - 8000x + 500$

Find the critical points by setting $C'(x) = 0$ :
$20x - 8000 = 0$ $20x = 8000$ $x = 400$