Is the following graph possible?

enter image description here

I've attempted to sketch the TC/TR curves for it but they don't seem to be able to satisfy the following two properties at the same time:

1) Slope of Total Cost (TC) being less than the slope of Total Revenue (TR) until q=b, that is until quantity is at the zero marginal profit level.

2) TC being lesser than TR at first but going on to be greater than it after q=a, that is after quantity is at the zero economic profit level.

  • $\begingroup$ Doesn't your graph imply both conditions (except perhaps TC being greater than TR close to zero, because revenue is $0$ when $q=0$, which FC $\not = 0$)? Slope of TC is MC, and slope of TR is MR, so 1) is true, and TC = AC *q, and price = D, so $2)$ is satisfied... $\endgroup$
    – majmun
    Jan 3, 2016 at 2:47
  • $\begingroup$ This question could be greatly improved by defining $a$ and $b$ in the text part of the question because search engines cannot search pictures and future users will be unable to find relevant info in questions such as these. $\endgroup$
    – Giskard
    Jan 3, 2016 at 7:07
  • $\begingroup$ Do you assume $F = 0$? Or do you mean $AVC$ instead of $AC$? $\endgroup$
    – Giskard
    Jan 7, 2016 at 10:58

1 Answer 1


Your basic assumptions seem to be $$ p(q_a) = AVC(q_a) $$ and as the difference between price and average variable cost is decreasing before $q_a$ and increasing after it you also assume $$ \frac{d \ \left(p(q) - AVC(q)\right)}{d \ q} < 0. $$ (More on the necessity of this assumption later.)
We can rewrite this second assumption to $$ \frac{d \ p(q)}{d \ q} - \frac{d \ \frac{VC(q)}{q}}{d \ q} = \frac{d \ p(q)}{d \ q} - \frac{MC(q) \cdot q - VC(q)}{q^2} < 0. $$ Multiplying by $q>0$ yields $$ \frac{d \ p(q)}{d \ q} \cdot q - MC(q) - AVC(q) < 0. $$ Adding $p(q) + AVC(q)$ we get $$ \frac{d \ p(q)}{d \ q} \cdot q + p(q) - MC(q) < p(q) - AVC(q). $$ The left hand side is now $MR(q) - MC(q)$. It follows from our two initial assumptions that $$ \forall q > q_a: \ p(q) < AVC(q). $$ Then for such values of $q$ $$ MR(q) - MC(q) < p(q) - AVC(q) < 0. $$ Because of this there can be no $q_b > q_a$ such that $MR(q_b) = MC(q_b)$.

On the necessity of the second assumption:
Seems to me that without this assumption the goal function $p(q) \cdot q - C(y)$ would not be concave and hence $MR(q) = MC(q)$ would not be a sufficient condition for optimum.


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