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The weekly demand for the Pulsar 25 color LED television is

[tex]\[ p = 600 - 0.05x \quad 0 \leq x \leq 12,000 \][/tex]

where [tex]\( p \)[/tex] denotes the unit price in dollars and [tex]\( x \)[/tex] denotes the quantity demanded. The weekly total cost function associated with manufacturing the Pulsar 25 is given by

[tex]\[ C(x) = 0.000002x^3 - 0.03x^2 + 400x + 80,000 \][/tex]

where [tex]\( C(x) \)[/tex] denotes the total cost incurred in producing [tex]\( x \)[/tex] sets.

Find the marginal profit [tex]\( P^{\prime}(x) \)[/tex]. Compute [tex]\( P^{\prime}(2000) \)[/tex] and interpret your results.

Sagot :

GinnyAnsweridn
To solve the problem, we need to find the marginal profit function [tex]\(P'(x)\)[/tex] and compute [tex]\(P'(2000)\)[/tex]. We'll also interpret the results of [tex]\(P'(2000)\)[/tex].

### Step-by-Step Solution:

1. Define the revenue function [tex]\(R(x)\)[/tex] and profit function [tex]\(P(x)\)[/tex]:

The revenue function [tex]\(R(x)\)[/tex] is found by multiplying the price [tex]\(p\)[/tex] by the quantity [tex]\(x\)[/tex]:
[tex]\[ R(x) = p \times x \][/tex]

Given the demand function:
[tex]\[ p = 600 - 0.05x \][/tex]

The revenue function becomes:
[tex]\[ R(x) = x(600 - 0.05x) \][/tex]
[tex]\[ R(x) = 600x - 0.05x^2 \][/tex]

The profit function [tex]\(P(x)\)[/tex] is the difference between the revenue [tex]\(R(x)\)[/tex] and the cost [tex]\(C(x)\)[/tex]:
[tex]\[ P(x) = R(x) - C(x) \][/tex]

Given the cost function:
[tex]\[ C(x) = 0.000002x^3 - 0.03x^2 + 400x + 80,000 \][/tex]

Substitute [tex]\(R(x)\)[/tex] and [tex]\(C(x)\)[/tex] into the profit function:
[tex]\[ P(x) = (600x - 0.05x^2) - (0.000002x^3 - 0.03x^2 + 400x + 80,000) \][/tex]

Simplify the expression:
[tex]\[ P(x) = 600x - 0.05x^2 - 0.000002x^3 + 0.03x^2 - 400x - 80,000 \][/tex]
[tex]\[ P(x) = -0.000002x^3 - 0.02x^2 + 200x - 80,000 \][/tex]

2. Find the marginal profit function [tex]\(P'(x)\)[/tex]:

The marginal profit is the derivative of the profit function [tex]\(P(x)\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ P'(x) = \frac{d}{dx} (-0.000002x^3 - 0.02x^2 + 200x - 80,000) \][/tex]

Differentiate each term:
[tex]\[ P'(x) = -0.000006x^2 - 0.04x + 200 \][/tex]

3. Compute [tex]\(P'(2000)\)[/tex]:

To find [tex]\(P'(2000)\)[/tex], substitute [tex]\(x = 2000\)[/tex] into the marginal profit function:
[tex]\[ P'(2000) = -0.000006(2000)^2 - 0.04(2000) + 200 \][/tex]
[tex]\[ P'(2000) = -0.000006(4000000) - 80 + 200 \][/tex]
[tex]\[ P'(2000) = -24 - 80 + 200 \][/tex]
[tex]\[ P'(2000) = 96 \][/tex]

### Interpretation of the result:

The marginal profit [tex]\(P'(2000) = 96\)[/tex] indicates that at a production level of 2000 units per week, the profit is increasing at a rate of [tex]$96 per additional unit produced. This means if the company increases production from 2000 units to 2001 units, the additional profit gained for that next unit is approximately $[/tex]96.
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