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Sagot :
Let's solve the given problem step-by-step.
### Given:
The cost function is [tex]\( C(x) = 250 \sqrt{x} + \frac{x^2}{125000} \)[/tex].
### a) The cost at the production level 1500
To find the cost at the production level of 1500, we simply substitute [tex]\( x = 1500 \)[/tex] into the cost function.
[tex]\[ C(1500) = 250 \sqrt{1500} + \frac{(1500)^2}{125000} \][/tex]
Calculating each term separately:
[tex]\[ 250 \sqrt{1500} \approx 250 \times 38.72983 \approx 9682.4575 \][/tex]
[tex]\[ \frac{(1500)^2}{125000} = \frac{2250000}{125000} = 18 \][/tex]
Therefore,
[tex]\[ C(1500) = 9682.4575 + 18 = 9700.458 \][/tex]
So, the cost at the production level 1500 is approximately [tex]\( 9700.458 \)[/tex].
### b) The average cost at the production level 1500
The average cost is given by [tex]\( \frac{C(x)}{x} \)[/tex]. Substituting [tex]\( x = 1500 \)[/tex]:
[tex]\[ \text{Average cost at } x = 1500 = \frac{C(1500)}{1500} = \frac{9700.458}{1500} \approx 6.467 \][/tex]
So, the average cost at the production level 1500 is approximately [tex]\( 6.467 \)[/tex].
### c) The marginal cost at the production level 1500
The marginal cost is the derivative of the cost function with respect to [tex]\( x \)[/tex].
[tex]\[ C(x) = 250 \sqrt{x} + \frac{x^2}{125000} \][/tex]
[tex]\[ \frac{dC}{dx} = 250 \cdot \frac{1}{2} x^{-1/2} + \frac{2x}{125000} \][/tex]
[tex]\[ = 125 x^{-1/2} + \frac{2x}{125000} \][/tex]
Now, evaluate this derivative at [tex]\( x = 1500 \)[/tex]:
[tex]\[ \left. \frac{dC}{dx} \right|_{x=1500} = 125 \cdot \frac{1}{\sqrt{1500}} + \frac{2 \times 1500}{125000} \][/tex]
[tex]\[ \approx 125 \cdot 0.02583 + 0.024 \][/tex]
[tex]\[ \approx 3.229 + 0.024 \][/tex]
[tex]\[ \approx 3.253 \][/tex]
So, the marginal cost at the production level 1500 is approximately [tex]\( 3.253 \)[/tex].
### d) The production level that will minimize the average cost
To find the production level that minimizes the average cost, we first need to find the average cost function.
[tex]\[ \text{Average cost} = \frac{C(x)}{x} = \frac{250 \sqrt{x} + \frac{x^2}{125000}}{x} = 250 \frac{\sqrt{x}}{x} + \frac{x^2}{125000 x} = \frac{250}{\sqrt{x}} + \frac{x}{125000} \][/tex]
Next, we find the derivative of this average cost function and set it to zero to find the critical points.
[tex]\[ \frac{d}{dx} \left( \frac{250}{\sqrt{x}} + \frac{x}{125000} \right) = -\frac{125}{x^{3/2}} + \frac{1}{125000} = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -\frac{125}{x^{3/2}} + \frac{1}{125000} = 0 \][/tex]
[tex]\[ \frac{125}{x^{3/2}} = \frac{1}{125000} \][/tex]
[tex]\[ 125 \cdot 125000 = x^{3/2} \][/tex]
[tex]\[ 15625000 = x^{3/2} \][/tex]
[tex]\[ x = \left( 15625000 \right)^{2/3} \][/tex]
[tex]\[ x = 62500 \][/tex]
So, the production level that will minimize the average cost is [tex]\( 62500 \)[/tex].
### e) The minimal average cost
Finally, we substitute [tex]\( x = 62500 \)[/tex] back into the average cost function to find the minimal average cost.
[tex]\[ \text{Minimal Average cost} = \frac{C(62500)}{62500} = \frac{250 \sqrt{62500} + \frac{(62500)^2}{125000}}{62500} \][/tex]
Calculate the terms:
[tex]\[ C(62500) = 250 \sqrt{62500} + \frac{(62500)^2}{125000} \][/tex]
[tex]\[ 250 \sqrt{62500} = 250 \times 250 = 62500 \][/tex]
[tex]\[ \frac{(62500)^2}{125000} = \frac{3906250000}{125000} = 31250 \][/tex]
So,
[tex]\[ C(62500) = 62500 + 31250 = 93750 \][/tex]
[tex]\[ \text{Minimal Average cost} = \frac{93750}{62500} = 1.5 \][/tex]
Thus, the minimal average cost is [tex]\( 1.5 \)[/tex].
### Summary of the Results:
- a) Cost at the production level 1500: [tex]\(9700.458\)[/tex]
- b) Average cost at the production level 1500: [tex]\(6.467\)[/tex]
- c) Marginal cost at the production level 1500: [tex]\(3.253\)[/tex]
- d) Production level that minimizes average cost: [tex]\(62500\)[/tex]
- e) Minimal average cost: [tex]\(1.5\)[/tex]
### Given:
The cost function is [tex]\( C(x) = 250 \sqrt{x} + \frac{x^2}{125000} \)[/tex].
### a) The cost at the production level 1500
To find the cost at the production level of 1500, we simply substitute [tex]\( x = 1500 \)[/tex] into the cost function.
[tex]\[ C(1500) = 250 \sqrt{1500} + \frac{(1500)^2}{125000} \][/tex]
Calculating each term separately:
[tex]\[ 250 \sqrt{1500} \approx 250 \times 38.72983 \approx 9682.4575 \][/tex]
[tex]\[ \frac{(1500)^2}{125000} = \frac{2250000}{125000} = 18 \][/tex]
Therefore,
[tex]\[ C(1500) = 9682.4575 + 18 = 9700.458 \][/tex]
So, the cost at the production level 1500 is approximately [tex]\( 9700.458 \)[/tex].
### b) The average cost at the production level 1500
The average cost is given by [tex]\( \frac{C(x)}{x} \)[/tex]. Substituting [tex]\( x = 1500 \)[/tex]:
[tex]\[ \text{Average cost at } x = 1500 = \frac{C(1500)}{1500} = \frac{9700.458}{1500} \approx 6.467 \][/tex]
So, the average cost at the production level 1500 is approximately [tex]\( 6.467 \)[/tex].
### c) The marginal cost at the production level 1500
The marginal cost is the derivative of the cost function with respect to [tex]\( x \)[/tex].
[tex]\[ C(x) = 250 \sqrt{x} + \frac{x^2}{125000} \][/tex]
[tex]\[ \frac{dC}{dx} = 250 \cdot \frac{1}{2} x^{-1/2} + \frac{2x}{125000} \][/tex]
[tex]\[ = 125 x^{-1/2} + \frac{2x}{125000} \][/tex]
Now, evaluate this derivative at [tex]\( x = 1500 \)[/tex]:
[tex]\[ \left. \frac{dC}{dx} \right|_{x=1500} = 125 \cdot \frac{1}{\sqrt{1500}} + \frac{2 \times 1500}{125000} \][/tex]
[tex]\[ \approx 125 \cdot 0.02583 + 0.024 \][/tex]
[tex]\[ \approx 3.229 + 0.024 \][/tex]
[tex]\[ \approx 3.253 \][/tex]
So, the marginal cost at the production level 1500 is approximately [tex]\( 3.253 \)[/tex].
### d) The production level that will minimize the average cost
To find the production level that minimizes the average cost, we first need to find the average cost function.
[tex]\[ \text{Average cost} = \frac{C(x)}{x} = \frac{250 \sqrt{x} + \frac{x^2}{125000}}{x} = 250 \frac{\sqrt{x}}{x} + \frac{x^2}{125000 x} = \frac{250}{\sqrt{x}} + \frac{x}{125000} \][/tex]
Next, we find the derivative of this average cost function and set it to zero to find the critical points.
[tex]\[ \frac{d}{dx} \left( \frac{250}{\sqrt{x}} + \frac{x}{125000} \right) = -\frac{125}{x^{3/2}} + \frac{1}{125000} = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -\frac{125}{x^{3/2}} + \frac{1}{125000} = 0 \][/tex]
[tex]\[ \frac{125}{x^{3/2}} = \frac{1}{125000} \][/tex]
[tex]\[ 125 \cdot 125000 = x^{3/2} \][/tex]
[tex]\[ 15625000 = x^{3/2} \][/tex]
[tex]\[ x = \left( 15625000 \right)^{2/3} \][/tex]
[tex]\[ x = 62500 \][/tex]
So, the production level that will minimize the average cost is [tex]\( 62500 \)[/tex].
### e) The minimal average cost
Finally, we substitute [tex]\( x = 62500 \)[/tex] back into the average cost function to find the minimal average cost.
[tex]\[ \text{Minimal Average cost} = \frac{C(62500)}{62500} = \frac{250 \sqrt{62500} + \frac{(62500)^2}{125000}}{62500} \][/tex]
Calculate the terms:
[tex]\[ C(62500) = 250 \sqrt{62500} + \frac{(62500)^2}{125000} \][/tex]
[tex]\[ 250 \sqrt{62500} = 250 \times 250 = 62500 \][/tex]
[tex]\[ \frac{(62500)^2}{125000} = \frac{3906250000}{125000} = 31250 \][/tex]
So,
[tex]\[ C(62500) = 62500 + 31250 = 93750 \][/tex]
[tex]\[ \text{Minimal Average cost} = \frac{93750}{62500} = 1.5 \][/tex]
Thus, the minimal average cost is [tex]\( 1.5 \)[/tex].
### Summary of the Results:
- a) Cost at the production level 1500: [tex]\(9700.458\)[/tex]
- b) Average cost at the production level 1500: [tex]\(6.467\)[/tex]
- c) Marginal cost at the production level 1500: [tex]\(3.253\)[/tex]
- d) Production level that minimizes average cost: [tex]\(62500\)[/tex]
- e) Minimal average cost: [tex]\(1.5\)[/tex]
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