A bakery’s profit satisfies MR = MC at two outputs: q = 3 and q = 14. At q = 3, π^′′ > 0; at q = 14, π^′′ < 0. Where should it produce?

At q = 3, the first crossing always wins

Wherever total revenue is highest

Business I / Optimization in Economics

Practice question · Multiple choice

A bakery’s profit satisfies $MR = MC$ at two outputs: $q = 3$ and $q = 14$ . At $q = 3$ , $\pi^{\prime\prime} > 0$ ; at $q = 14$ , $\pi^{\prime\prime} < 0$ . Where should it produce?

Hints

$MR = MC$ holds at minima too.
Positive curvature at $q=3$ means a valley.

Show the answer

C. At $q = 14$ , confirmed by the SOC

Why

$q = 3$ is a profit minimum dressed in the same FOC; $q = 14$ passes the second-order test. This is precisely why the ritual has two steps, the first equation alone cannot tell a peak from a pit.

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A bakery’s profit satisfies MR=MCMR = MCMR=MC at two outputs: q=3q = 3q=3 and q=14q = 14q=14. At q=3q = 3q=3, π′′>0\pi^{\prime\prime} > 0π′′>0; at q=14q = 14q=14, π′′<0\pi^{\prime\prime} < 0π′′<0. Where should it produce?

Practise Optimization in Economics

More questions on Optimization in Economics

A bakery’s profit satisfies $MR = MC$ at two outputs: $q = 3$ and $q = 14$ . At $q = 3$ , $\pi^{\prime\prime} > 0$ ; at $q = 14$ , $\pi^{\prime\prime} < 0$ . Where should it produce?