HEC Montréal MATH10600 tutoring

1. Write the profit function. Let $P(q)$ denote profit in dollars. It is revenue minus cost, so $P(q)=R(q)-C(q)=-200+30q-0.5q^2$. The fixed cost is still present at zero output in this model. 2. Find a possible maximum inside the allowed interval. A prime denotes a derivative with respect to output. The rate of change of profit, called marginal profit, is $P'(q)=30-q$, in dollars per unit. Setting $P'(q)=0$ gives $q=30$, which lies in the feasible interval. The second derivative is $P''(q)=-1$, so the graph is strictly concave: it bends downward everywhere. 3. Check the whole domain. Because profit is continuous on the closed interval, compare the profit at the candidate output with the profit at both endpoints: $P(0)=-200$, $P(30)=-200+900-450=250$, and $P(60)=-200$, all in dollars. Therefore the maximum profit is $250$ dollars at $30$ units. At this output, marginal revenue $R'(q)=40$ equals marginal cost $C'(q)=10+q=40$, both in dollars per unit. 4. Apply the new constraint. With capacity $20$, the domain becomes $[0,20]$, so the former candidate is unavailable. On this whole interval, $P'(q)=30-q\ge10>0$, meaning profit rises as output rises. The maximum is therefore at the boundary $q=20$, where $P(20)=-200+600-200=200$ dollars. A zero derivative identifies a possible maximum inside the interval. You must still check that the output is allowed and compare the profit at the endpoints.