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Answer any one of the following.
The Average Cost of producing 'x' units of commodity is given by: AC = x/(200) + 30 + (5000)/x Find…
The Average Cost of producing '$x$' units of commodity is given by:
$AC = \frac{x}{200} + 30 + \frac{5000}{x}$
(a)[1]
Find the Cost function.
(b)[1]
Find the Marginal Cost function.
(c)[1]
Find the Marginal Average Cost function.
(d)[1]
Verify that $\frac{d}{dx}(AC) = \frac{MC - AC}{x}$
Given: AC = \frac{x}{200} + 30 + \frac{5000}{x}
To show: \frac{d}{dx}(AC) = \frac{MC - AC}{x}
Answer
Answer (a)
$C(x)=x\cdot AC=\frac{x^2}{200}+30x+5000$.
Final answer: $C(x)=\frac{x^2}{200}+30x+5000$
Answer (b)
$MC=\frac{dC}{dx}=\frac{x}{100}+30$.
Final answer: $MC=\frac{x}{100}+30$
Answer (c)
Differentiating: x/200 + 30 + 5000/x
Marginal Average Cost $=\frac{d}{dx}(AC)=\frac1{200}-\frac{5000}{x^2}$.
Final answer: $\frac1{200}-\frac{5000}{x^2}$
Answer (d)
Verification that $\frac{d}{dx}(AC)=\frac{MC-AC}{x}$
- 1. $\frac{d}{dx}(AC)=\frac1{200}-\frac{5000}{x^2}$.
- 2. $MC-AC=\frac x{100}+30-\frac x{200}-30-\frac{5000}x=\frac x{200}-\frac{5000}x$.
- 3. $\frac{MC-AC}{x}=\frac1{200}-\frac{5000}{x^2}$.
- 4. Hence $\frac{d}{dx}(AC)=\frac{MC-AC}{x}$.