Let's solve the given problem step-by-step given the cost function [tex]\(C(x) = 57600 + 800x + x^2\)[/tex]:
### a) The cost at the production level 1300
To find the cost at a production level of 1300 units, we substitute [tex]\(x = 1300\)[/tex] into the cost function [tex]\(C(x)\)[/tex].
[tex]\[ C(1300) = 57600 + 800(1300) + (1300)^2 \][/tex]
Substituting [tex]\(x = 1300\)[/tex]:
[tex]\[ C(1300) = 57600 + 1040000 + 1690000 = 2787600 \][/tex]
So, the cost at the production level of 1300 units is 2787600.
### b) The average cost at the production level 1300
The average cost function is given by:
[tex]\[ \text{Average Cost} = \frac{C(x)}{x} \][/tex]
At [tex]\( x = 1300 \)[/tex]:
[tex]\[ \text{Average Cost at 1300} = \frac{C(1300)}{1300} = \frac{2787600}{1300} = \frac{27876}{13} \][/tex]
So, the average cost at the production level of 1300 units is [tex]\(\frac{27876}{13}\)[/tex].
### c) The marginal cost at the production level 1300
The marginal cost is the derivative of the cost function [tex]\(C(x)\)[/tex] with respect to [tex]\(x\)[/tex]. We first find this derivative and then substitute [tex]\(x = 1300\)[/tex]:
[tex]\[ C(x) = 57600 + 800x + x^2 \][/tex]
[tex]\[ C'(x) = 800 + 2x \][/tex]
At [tex]\( x = 1300 \)[/tex]:
[tex]\[ C'(1300) = 800 + 2(1300) = 800 + 2600 = 3400 \][/tex]
So, the marginal cost at the production level of 1300 units is 3400.
### d) The production level that will minimize the average cost
To minimize the average cost, we need to work with the average cost function:
[tex]\[ \text{Average Cost} = \frac{C(x)}{x} = \frac{57600 + 800x + x^2}{x} = \frac{57600}{x} + 800 + x \][/tex]
We need to find the critical points by taking the derivative of the average cost function and setting it to zero:
[tex]\[ \text{Average Cost Prime} = -\frac{57600}{x^2} + 1 \][/tex]
Setting it to zero to find the critical points:
[tex]\[ -\frac{57600}{x^2} + 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{57600}{x^2} = 1 \][/tex]
[tex]\[ x^2 = 57600 \][/tex]
[tex]\[ x = \sqrt{57600} = 240 \][/tex]
So, the production level that will minimize the average cost is 240 units.
### e) The minimal average cost
Finally, to find the minimal average cost, we substitute [tex]\( x = 240 \)[/tex] into the average cost function:
[tex]\[ \text{Minimal Average Cost} = \frac{C(240)}{240} \][/tex]
First, find [tex]\(C(240)\)[/tex]:
[tex]\[ C(240) = 57600 + 800(240) + (240)^2 = 57600 + 192000 + 57600 = 307200 \][/tex]
Now, calculate the average cost:
[tex]\[ \text{Minimal Average Cost} = \frac{307200}{240} = 320 \][/tex]
So, the minimal average cost is 320.
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In summary:
a) The cost at the production level 1300 is 2787600.
b) The average cost at the production level 1300 is [tex]\(\frac{27876}{13}\)[/tex].
c) The marginal cost at the production level 1300 is 3400.
d) The production level that will minimize the average cost is 240.
e) The minimal average cost is 320.
