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A firm's average cost function is: $$AC = x^2 - 2x + 50 , \ 0 \leq x$$ a) Find the level of output to minimalize average cost (AC) function

b) Find the level of output when average cost equals marginal cost (AC=MC)

Could you please check my calculations ? I have done a) and a little b):

a)$$AC = x^2 - 2x + 50$$ $$f'(x)=2x-2=0$$ $$x-1=0$$ $$x=1$$ $$f(1) = 1 - 2 + 50 = 49$$

b) This one I'm totally not sure $$TC=AC·x$$ $$TC=x^3 - 2x^2 + 50x$$ $$MC=TC'=3x^2-4x+50$$ $$AC=MC$$ $$x^2 - 2x + 50=3x+2-4x+50$$ $$x_1=1$$$$x_2=0$$

Michael
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  • Isn't marginal cost the derivative of total cost? – lulu May 31 '16 at 11:54
  • It would help if you explained some of the steps, although I could guess some of them (for instance in a) where you differentiate and what not). If you want to express symbolically, then be precise and clear.

    a) Seems correct to me (if $x$ is level of output). You might want to add that $AC \to \infty$ when $x \to \pm \infty$, so your point is not only a local minimum.

    b) Furthermore, how do you define marginal cost? Is it defined as the integral of $TC$? Are you sure that there are not bounds for the integral, otherwise it is not solvable.

     –  May 31 '16 at 11:54
  • I'm not sure what how to solve b) but @lulu said that marginal cost is the derivative of total cost and I think using this is possible to solve it – Michael May 31 '16 at 12:17
  • I edited the solution, could you say what to do now ? – Michael May 31 '16 at 12:26
  • Your solution is correct. – Emre May 31 '16 at 12:27
  • So both x1 and x2 are solutions of the task? – Michael May 31 '16 at 12:31
  • As a check, turning points of the average cost occur when it is equal to the marginal cost. One way to show something is a local minimum (after showing that its first derivative is zero) is to show its second derivative is positive there. – Henry May 31 '16 at 12:40

1 Answers1

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The average cost function is the parabola for $x\ge 0$ $$\operatorname{AC}(x)=(x-1)^2+49$$ which has the minimum at the point $P=(1,49)$.

The total cost function is $\operatorname{TC}(x)=\operatorname{AC}(x)\cdot x$ and then differentiating $$ \underbrace{\operatorname{TC}'(x)}_{\operatorname{MC}(x)}=\operatorname{AC}'(x)\cdot x+\operatorname{AC}(x) $$ and using $\operatorname{MC}(x)=\operatorname{AC}(x)$ and $\operatorname{AC}'(x)=2(x-1)$ we have $$ 0=\operatorname{AC}'(x)\cdot x=2(x-1)x $$ that is $x=0$ or $x=1$.

alexjo
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