Marianopolis College · 201-SH2-MS

Marianopolis 201-SH2-MS: Differential Calculus in Social Science Tutoring

A derivative can describe a slope, a rate of change or the approximate effect of producing one more unit. In 201-SH2-MS, interpreting that result is as important as calculating it. Nablio helps Marianopolis students move between formulas, graphs and business or social-science questions. Start with the profit example below to see why maximizing revenue and maximizing profit can give different answers.

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Use derivatives with a clear interpretation

Limits and continuity: identify what a function approaches and whether its value agrees at the point.
Derivative definitions and rules: connect a difference quotient to an instantaneous rate before simplifying.
Graphs and optimization: use signs, critical points and the allowed domain to justify a conclusion.
Marginal analysis: distinguish total, average and marginal cost or revenue, with units attached.

Write the quantity you want to optimize

Define the decision variable and its feasible interval. If the goal is profit, form revenue minus cost before differentiating; setting marginal revenue to zero solves a different problem. After finding a candidate, check whether it is a maximum on the permitted interval. In a continuous model, a marginal value estimates the change from a small increase in output; it is not generally the exact cost of the next whole unit.

Worked example: choose output to maximize profit

In an original continuous-output model, selling q units gives a price of $p(q)=100-0.5q$ dollars per unit. Total cost is $C(q)=400+20q$ dollars, with $0\le q\le200$ . Find the output, selling price and profit at the maximum.

Revenue is price times quantity: $R(q)=q(100-0.5q)$ . Subtract total cost to obtain $P(q)=80q-0.5q^2-400$ .
Differentiate the profit: $P'(q)=80-q$ . The only stationary point is $q=80$ , which lies inside the allowed interval.
Since $P''(q)=-1<0$ throughout the interval, the profit function is strictly concave and this point is its unique maximum.
The price is $p(80)=60$ dollars per unit. Revenue is 4,800 dollars and cost is 2,000 dollars, giving a maximum profit of 2,800 dollars.
Check the economics: $R'(80)=100-80=20$ and $C'(80)=20$ dollars per unit. Marginal revenue equals marginal cost at the maximizing output.

Maximizing revenue alone would give q = 100. Once variable costs are included, the profit-maximizing output is q = 80. A zero derivative needs both a clearly defined objective and a maximum check.

What if the variable cost increases?

Keep the same price function and feasible interval, but change cost to $C(q)=400+30q$ . Find the new maximizing output, price and profit. Explain the direction of the output change.

Show the explained answer

Profit is $P(q)=70q-0.5q^2-400$ , so $P'(q)=70-q$ gives q = 70. The second derivative is still negative. Price is 65 dollars per unit and profit is 2,050 dollars. Higher marginal cost makes the previous last units less profitable, so the optimal output decreases.

Original Nablio example and practice, independent of the college. These are not course assessment questions. Follow your current section’s outline for the material to review.

Choose the next check from your attempt

If the profit function was right but its derivative was wrong, use the guided derivative-rule check. Name the outermost operation before differentiating; the later questions combine several rules.
If total cost, average cost and marginal cost became interchangeable, use marginal cost as a rate. Then compare a local estimate with the exact change. These first two sections continue the differential-calculus work; the later integration section is a separate next step.

For each new model, identify the quantity to optimize and the feasible interval, and check whether output is treated as continuous. Use that model’s assumptions when setting up its calculation.

Work on the setup as well as the derivative

Bring your current course outline and the original wording of a question, if you have them. You can start the question with your tutor or review any work you have tried. A session can focus on building a function from the context, applying the appropriate derivative rule or interpreting a sign chart. The aim is to make your reasoning clear enough to use on a new problem.

We use permitted practice without completing graded assignments or exams on your behalf.

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