For the given cost function [tex]C(x)=250 \sqrt{x}+\frac{x^2}{125000}[/tex], find:

a) The cost at the production level 1500.
b) The average cost at the production level 1500.
c) The marginal cost at the production level 1500.
d) The production level that will minimize the average cost.
e) The minimal average cost.

Give answers to at least 3 decimal places.



Answer :

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]

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