Profit, domain restrictions and an integer decision
A fictional workshop models the revenue from a batch of q items by R(q)=60q-q² and its total cost by C(q)=100+10q, using consistent currency units. The workshop can make at most40 items in a batch. Its planner first treats q as continuous on0≤q≤40 to locate a maximum, then returns to whole-number production quantities. The fixed cost is incurred even for a batch of zero items. The model is supplied for a mathematics exercise and does not represent a recommendation about an actual business or an investment. A colleague suggests maximizing revenue alone. Another suggests solving for the point where profit is zero. The planner needs a distinction between those questions and maximizing profit, including the effect of both endpoints.
1. Form the profit function and find its stationary quantity.
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P(q)=R(q)-C(q)=-q²+50q-100. Its derivative is P′(q)=-2q+50, so the stationary quantity is q=25. Revenue alone is a different function and would not account for production costs.
2. Check the domain and endpoints, then state the profit-maximizing integer quantity.
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P″=-2<0, so the stationary point is a maximum. It lies in[0,40]. P(0)=-100, P(40)=300 and P(25)=525. Thus the admissible integer quantity25 gives the maximum modelled profit525.
3. Find the least positive integer quantity giving positive profit.
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P(q)>0 holds between the roots25±5√21, approximately2.087 and47.913. Intersect this interval with[0,40]. The least integer is3; P(2)=-4 while P(3)=41.