Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
