John Abbott · Social Science · 201-SH2-AB

201-SH2-AB Calculus 1 tutoring for Social Science

Get one-to-one help with limits, derivatives and optimization for John Abbott 201-SH2-AB. Work through a short business example below, or bring a question to a tutor, even if you have not started it.

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Calculus for Social Science and Commerce

John Abbott’s Calculus 1 for Social Science connects functions, limits and derivatives with curve sketching, optimization and business applications. You also work with exponential, logarithmic and trigonometric functions. Start with the calculation or interpretation that is slowing you down.

Topics to work on

Algebra, functions, limits and continuity
Derivative rules for algebraic, exponential, logarithmic and trigonometric functions
Curve sketching and the meaning of a derivative’s sign
Optimization and business applications

A useful starting point

Review expanding brackets, function notation and reading an interval. In the example, you will turn a price and a cost into a profit function, differentiate a quadratic, and compare the permitted quantities.

Steps worth checking

Distinguishing price per item, total revenue and total profit
Checking the allowed domain before accepting a critical point

Worked example: choose a quantity to maximize profit

In this simplified sales model, let $q$ be the number of items sold, treated as a continuous variable on $0\le q\le80$ . All items sell at the price $p(q)=50-\tfrac12q$ dollars per item. The total cost is $C(q)=100+10q$ dollars. The fixed cost of 100 dollars applies even when no items are sold. Find the quantity that gives the largest profit, and state the corresponding price and profit.

Show a hint

First multiply the price per item by the quantity to get total revenue. Subtract the total cost, including the fixed cost. A zero derivative gives a candidate; check whether it is allowed and whether it really gives the largest profit.

Show the worked solution

1. Build the profit function
Let $R(q)$ be total revenue and $P(q)$ be total profit, both in dollars. Revenue is price times quantity, and profit is revenue minus cost:
$\begin{aligned}R(q)&=q\left(50-\tfrac12q\right),\\P(q)&=R(q)-C(q)\\&=40q-\tfrac12q^2-100.\end{aligned}$
2. Find and justify the maximum
The derivative $P^{\prime}(q)=40-q$ is the instantaneous rate of change of profit with respect to quantity, measured in dollars per item. It is zero at $q=40$ , positive for $0<q<40$ , and negative for $40<q<80$ . Profit therefore increases up to 40 items and decreases afterward.
The function is continuous on the closed interval $[0,80]$ . Check the interior candidate and the endpoints: $P(0)=-100,\quad P(40)=700,\quad P(80)=-100.$ The largest profit is 700 dollars at 40 items. This quantity is also a whole number, so it is feasible if items must be sold individually.
3. Interpret the result and check the money
At $q=40$ , the price is $p(40)=30$ dollars per item. Revenue is $40\times30=1200$ dollars and cost is $100+10\times40=500$ dollars. Their difference is the profit: $1200-500=700$ dollars.
The answer is a quantity of 40 items, a price of 30 dollars per item, and a profit of 700 dollars. The derivative’s value of zero is a rate, not the maximum profit.

Your turn: a smaller capacity, then a higher fixed cost

Use the same price and cost model, but the seller can now sell at most 30 items. (1) Write the new interval for $q$ . (2) Find the quantity, price and largest profit on that interval. (3) Keeping this 30-item limit, suppose the fixed cost rises from 100 to 160 dollars. Does the best quantity change? What is the new maximum profit?

Show a hint

The formula $P^{\prime}(q)=40-q$ has not changed in part (2), but the allowed interval has. For part (3), write the new profit as the old profit minus a constant.

Check the three steps

1. Keep only feasible quantities
The new interval is $0\le q\le30$ . The original candidate $q=40$ is outside it and cannot be chosen.
2. Use the sign on the new interval
For $0<q<30$ , $P^{\prime}(q)=40-q>0$ , so profit increases throughout the permitted interval. The maximum is at the endpoint $q=30$ .
The price is $p(30)=35$ dollars per item. Revenue is $30\times35=1050$ dollars, cost is $100+10\times30=400$ dollars, and profit is $1050-400=650$ dollars. This is greater than $P(0)=-100$ . An endpoint maximum does not require a zero derivative.
3. Separate the amount of profit from the best quantity
Call the new profit $\widetilde P(q)$ . Then $\widetilde P(q)=P(q)-60$ , so $\widetilde P^{\prime}(q)=40-q$ . Every profit value falls by the same 60 dollars; the maximizing quantity stays at 30 items and the price stays at 35 dollars per item. The new maximum profit is $650-60=590$ dollars.

If building the function was difficult, write the units beside price, revenue and cost. If the derivative was right but the conclusion was unclear, explain the sign change and check the endpoints before trying the self-check.

What to work through with a tutor

Bring a question, even if you do not know how to start. For a limit, your tutor can check the domain and the algebra before you substitute. For a derivative, they can help you choose the rule, keep every factor and check the result.

For a business application, your tutor can help you define the quantity and its units, build revenue or profit from the wording, and justify an optimum using the derivative and the domain. You can then try a changed version yourself and explain what stays the same.

Build your review from your course outline

Bring the topics your teacher has announced and a practice question, with any work you have started. Mix a limit, a derivative and an application from that scope. Explain the method before calculating, then redo a missed question without the solution. Follow your section’s instructions for permitted tools and assessment requirements.

Independent practice

Choose your next practice step

30 min

Before Calculus: 10 Algebra Skills You Should Know

Review fractions, factoring and exponents when algebra is interrupting the calculus.

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Derivative Rules: A Three-Question Check

Choose the product, quotient or chain rule, then check each factor.

14 questions

Calculus I Practice for Limits and Derivatives

Select questions on limits, continuity and derivative methods that match your current topics.

Course information

John Abbott: Social Science course descriptions

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