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Sagot :
Sure, let's go through each part of solving this problem step-by-step:
### Given the cost function [tex]\( C(x) = 12100 + 800x + x^2 \)[/tex]
#### a) The cost at the production level 1650
To find the cost at the production level [tex]\( x = 1650 \)[/tex], we simply substitute [tex]\( x \)[/tex] with 1650 in the cost function:
[tex]\[ C(1650) = 12100 + 800 \times 1650 + 1650^2 \][/tex]
After calculating, we find:
[tex]\[ C(1650) = 4,054,600 \][/tex]
#### b) The average cost at the production level 1650
The average cost [tex]\( AC \)[/tex] at a production level is given by dividing the total cost [tex]\( C(x) \)[/tex] by the number of units produced [tex]\( x \)[/tex]:
[tex]\[ AC(1650) = \frac{C(1650)}{1650} \][/tex]
From part (a), we know [tex]\( C(1650) = 4,054,600 \)[/tex], so:
[tex]\[ AC(1650) = \frac{4,054,600}{1650} = \frac{7372}{3} \][/tex]
#### c) The marginal cost at the production level 1650
The marginal cost [tex]\( MC \)[/tex] is the derivative of the cost function with respect to [tex]\( x \)[/tex], evaluated at [tex]\( x = 1650 \)[/tex]:
[tex]\[ C(x) = 12100 + 800x + x^2 \][/tex]
Taking the derivative:
[tex]\[ C'(x) = 800 + 2x \][/tex]
Evaluating at [tex]\( x = 1650 \)[/tex]:
[tex]\[ MC(1650) = 800 + 2 \times 1650 = 4100 \][/tex]
#### d) The production level that will minimize the average cost
To minimize the average cost, we first need the average cost function [tex]\( AC(x) \)[/tex]:
[tex]\[ AC(x) = \frac{C(x)}{x} = \frac{12100 + 800x + x^2}{x} \][/tex]
Simplifying, we get:
[tex]\[ AC(x) = \frac{12100}{x} + 800 + x \][/tex]
To find the minimum, we take the derivative with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{d}{dx} \left( \frac{12100}{x} + 800 + x \right) = -\frac{12100}{x^2} + 1 \][/tex]
Setting the derivative equal to zero:
[tex]\[ -\frac{12100}{x^2} + 1 = 0 \quad \Rightarrow \quad \frac{12100}{x^2} = 1 \quad \Rightarrow \quad x^2 = 12100 \quad \Rightarrow \quad x = -\sqrt{12100} \quad (\text{Ignoring } \sqrt{12100} \text{ since } x > 0) \][/tex]
[tex]\[ x = -110 \quad (\text{approximating } \sqrt{12100} \text{ as a negative root}) \][/tex]
#### e) The minimal average cost
Substituting [tex]\( x \)[/tex] from part (d) back into the average cost function:
[tex]\[ AC(-110) = \frac{12100}{-110} + 800 - 110 \][/tex]
Simplifying:
[tex]\[ AC(-110) = -110 + 800 - 110 = 580 \][/tex]
So the minimal average cost is:
[tex]\[ 580 \][/tex]
These steps give us the detailed solution for each part of the problem.
### Given the cost function [tex]\( C(x) = 12100 + 800x + x^2 \)[/tex]
#### a) The cost at the production level 1650
To find the cost at the production level [tex]\( x = 1650 \)[/tex], we simply substitute [tex]\( x \)[/tex] with 1650 in the cost function:
[tex]\[ C(1650) = 12100 + 800 \times 1650 + 1650^2 \][/tex]
After calculating, we find:
[tex]\[ C(1650) = 4,054,600 \][/tex]
#### b) The average cost at the production level 1650
The average cost [tex]\( AC \)[/tex] at a production level is given by dividing the total cost [tex]\( C(x) \)[/tex] by the number of units produced [tex]\( x \)[/tex]:
[tex]\[ AC(1650) = \frac{C(1650)}{1650} \][/tex]
From part (a), we know [tex]\( C(1650) = 4,054,600 \)[/tex], so:
[tex]\[ AC(1650) = \frac{4,054,600}{1650} = \frac{7372}{3} \][/tex]
#### c) The marginal cost at the production level 1650
The marginal cost [tex]\( MC \)[/tex] is the derivative of the cost function with respect to [tex]\( x \)[/tex], evaluated at [tex]\( x = 1650 \)[/tex]:
[tex]\[ C(x) = 12100 + 800x + x^2 \][/tex]
Taking the derivative:
[tex]\[ C'(x) = 800 + 2x \][/tex]
Evaluating at [tex]\( x = 1650 \)[/tex]:
[tex]\[ MC(1650) = 800 + 2 \times 1650 = 4100 \][/tex]
#### d) The production level that will minimize the average cost
To minimize the average cost, we first need the average cost function [tex]\( AC(x) \)[/tex]:
[tex]\[ AC(x) = \frac{C(x)}{x} = \frac{12100 + 800x + x^2}{x} \][/tex]
Simplifying, we get:
[tex]\[ AC(x) = \frac{12100}{x} + 800 + x \][/tex]
To find the minimum, we take the derivative with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{d}{dx} \left( \frac{12100}{x} + 800 + x \right) = -\frac{12100}{x^2} + 1 \][/tex]
Setting the derivative equal to zero:
[tex]\[ -\frac{12100}{x^2} + 1 = 0 \quad \Rightarrow \quad \frac{12100}{x^2} = 1 \quad \Rightarrow \quad x^2 = 12100 \quad \Rightarrow \quad x = -\sqrt{12100} \quad (\text{Ignoring } \sqrt{12100} \text{ since } x > 0) \][/tex]
[tex]\[ x = -110 \quad (\text{approximating } \sqrt{12100} \text{ as a negative root}) \][/tex]
#### e) The minimal average cost
Substituting [tex]\( x \)[/tex] from part (d) back into the average cost function:
[tex]\[ AC(-110) = \frac{12100}{-110} + 800 - 110 \][/tex]
Simplifying:
[tex]\[ AC(-110) = -110 + 800 - 110 = 580 \][/tex]
So the minimal average cost is:
[tex]\[ 580 \][/tex]
These steps give us the detailed solution for each part of the problem.
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