# Is the following graph possible?

Is the following graph possible?

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.

• 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... – majmun Jan 3 '16 at 2:47
• 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. – Giskard Jan 3 '16 at 7:07
• Do you assume $F = 0$? Or do you mean $AVC$ instead of $AC$? – Giskard Jan 7 '16 at 10:58

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 $MR(q_b) = MC(q_b)$.
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.