Dawson 201-MA1-DW: Social Science Calculus I Tutoring

Listed course

Course overview

Dawson’s 201-MA1-DW connects limits and derivatives with Social Science applications, including marginal analysis and optimization. Work through the cost example below to distinguish an average rate of change from a marginal rate, then build and maximize a profit function.

201-MA1-DW topics to review

Limits and continuityDerivatives and curve sketchingMarginal analysisOptimization

Common difficulties

Keeping total cost, average rate of change and marginal cost distinct, with the right units
Building a profit function and justifying its maximum over the allowed quantities

Preparing for this course

Use this practice to explain what a derivative tells you about a cost model, then use it to choose a production quantity. Start by substituting into a quadratic, expanding brackets and solving a linear equation. Read the worked example in order: total cost, average rate of change, marginal cost, then an exact change. Try the profit problem with the solution closed. A prime, as in $C^\prime(q)$ , means a derivative with respect to $q$ .

Registration requirements

Dawson’s course list states that 201-MA1-DW cannot replace 201-NYA-05. Confirm the registration requirements for your program with Dawson.

Check the official course catalogue

From a cost model to marginal cost

Original practice. A workshop models its total cost for one day by $C(q)=200+8q+0.02q^2$ , in Canadian dollars, where $q$ is the number of items produced that day and $0\le q\le100$ . Treat $q$ as continuous when differentiating; actual production uses whole items. First find $C(40)$ and the average rate of change in cost from 40 to 45 items. Next find $C^\prime(40)$ and explain its units. Use it to estimate the additional cost of increasing production from 40 to 41 items, then calculate that additional cost exactly. Why do the estimate and exact change differ?

Show a hint

An average rate uses a change in cost divided by a change in quantity: $[C(45)-C(40)]/(45-40)$ . For the derivative, apply the power rule to each term. To estimate a cost change, multiply the marginal cost by the change in quantity. Keep that estimate separate from the subtraction $C(41)-C(40)$ .

Show the reasoning

Read the cost values
The constant term gives $C(0)=200$ : the model includes CAD 200 of cost even when no items are produced. At 40 items, $C(40)=200+8(40)+0.02(40)^2=552$ . The total cost for that day is CAD 552.
At 45 items, $C(45)=200+360+40.50=600.50$ . These outputs are amounts of money; neither is a rate.
Calculate an average rate over five additional items
$\frac{C(45)-C(40)}{45-40}=\frac{600.50-552}{5}=9.70.$
The extra five items add CAD 48.50 to total cost, an average increase of CAD 9.70 per additional item. On a cost graph, this is the slope of the secant joining the two cost points. It is different from dividing total cost by total output, which would give the average cost per item at one production level.
Find the marginal cost at 40 items
The derivative of the constant is zero, the derivative of $8q$ is 8, and the derivative of $0.02q^2$ is $0.04q$ . Thus $C^\prime(q)=8+0.04q$ , so $C^\prime(40)=9.60$ . Its unit is Canadian dollars per item.
This derivative is the tangent slope at 40 items. It tells us the local rate at which the model’s cost changes as quantity increases. It supplies an estimate of a small cost change once we multiply by a change in quantity.
Connect the derivative to a shrinking change in quantity
Let $h\ne0$ be a change in the number of items, with $40+h$ still between 0 and 100. Expanding the cost model gives $C(40+h)-C(40)=9.60h+0.02h^2$ . Therefore
$\frac{C(40+h)-C(40)}{h}=9.60+0.02h.$
As $h$ approaches zero, this average rate approaches 9.60, agreeing with the derivative. At $h=5$ , the same expression gives 9.70, agreeing with the five-item calculation. The continuous model lets us take this limit even though actual items are counted in whole numbers.
Compare the estimate with the exact next-item cost
For one additional item, the derivative estimate is $C^\prime(40)\times1=9.60$ Canadian dollars. The exact calculation gives $C(41)=200+328+33.62=561.62$ , so $C(41)-C(40)=9.62$ Canadian dollars.
The estimate is two cents lower. In the expansion above, the linear estimate keeps $9.60h$ and leaves out $0.02h^2$ , which is 0.02 at $h=1$ . The marginal cost also increases with quantity, so the starting tangent slope understates the cost increase over this positive interval. State both results clearly: CAD 9.60 is the estimate; CAD 9.62 is the exact increase under this model.

Your turn: choose the quantity that maximizes profit

Keep the same cost function and the allowed quantities $0\le q\le100$ . Suppose all items are sold at a price $p(q)=20-0.10q$ Canadian dollars per item. (1) Write the revenue $R(q)$ and the profit $\Pi(q)$ , both in Canadian dollars for the day. (2) Check the profit at 40 items. (3) Find the quantity that gives the greatest profit over the whole allowed interval, justify the maximum, and report the price and profit there. Start by multiplying price per item by the number sold, then subtract the total cost.

Check your reasoning

Build the quantity you want to maximize
Revenue is price multiplied by quantity: $R(q)=q(20-0.10q)=20q-0.10q^2$ . Profit is revenue minus total cost:
$\Pi(q)=-0.12q^2+12q-200.$
At 40 items, $p(40)=16$ , $R(40)=640$ , and $C(40)=552$ . Profit is therefore $640-552=88$ Canadian dollars. Keeping the fixed cost of 200 in the subtraction matters for the final profit.
Find and justify the best quantity
Differentiate: $\Pi^\prime(q)=12-0.24q$ . Setting the derivative to zero gives $q=50$ , which lies in the allowed interval. The derivative is positive for $0<q<50$ and negative for $50<q<100$ . Profit increases up to 50 and decreases afterward, so 50 gives the global maximum.
The endpoint profits are $\Pi(0)=-200$ and $\Pi(100)=-200$ , while $\Pi(50)=100$ . These values confirm the comparison on the closed interval. Simply solving $\Pi^\prime(q)=0$ would not, by itself, justify a maximum.
Return to price, cost and profit
At 50 items, $p(50)=15$ Canadian dollars per item. Revenue is $R(50)=750$ and cost is $C(50)=650$ , so the maximum daily profit is CAD 100. The maximizing quantity is already a whole number, so it also solves the whole-item production problem.
As a check on the derivative reasoning, $R^\prime(50)=10=C^\prime(50)$ : marginal revenue equals marginal cost at this interior optimum. The derivative sign change and the allowed interval establish why that equality gives the maximum here.

Compare marginal cost with an exact cost change Start with the first two sections for total cost, average unit cost and a marginal estimate. The lesson uses a different cost model, so read its coefficients and domain before calculating.
Check a derivative rule Use this if the economic model is clear but the derivative step is not.

How tutoring can help

Bring a cost or profit question, even if you do not know which function to differentiate. Your tutor can help you choose a first step or review a calculation you have started, whether the difficulty is expansion, the derivative rule or the economic meaning of the function. You can compare a secant with a tangent, explain the dollars-per-item units, and then build a profit function together. Finish by changing a coefficient and justifying the new maximum yourself.

Exam preparation for 201-MA1-DW

Build a short practice set in stages: evaluate a function, calculate an average rate, interpret a derivative, then solve an optimization problem. For each application, state the variable, units and allowed interval before calculating. On a second attempt, cover the worked solution and explain why the maximum is valid. Add limits, continuity or curve-sketching questions from the chapters you are reviewing.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Useful links for preparation or further study—not transfer equivalencies or a required registration sequence.

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201-MA1-DW · official course reference 201-MA1-DW · official offering information

Dawson 201-MA1-DW tutoring

Course overview

201-MA1-DW topics to review

Common difficulties

Preparing for this course

Registration requirements

From a cost model to marginal cost

Read the cost values

Calculate an average rate over five additional items

Find the marginal cost at 40 items

Connect the derivative to a shrinking change in quantity

Compare the estimate with the exact next-item cost

Your turn: choose the quantity that maximizes profit

Build the quantity you want to maximize

Find and justify the best quantity

Return to price, cost and profit

Choose your next practice step

How tutoring can help

Exam preparation for 201-MA1-DW

Related courses

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