Concordia University · MATH 208

Concordia MATH 208: Fundamental Mathematics I tutoring

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Course overview

Concordia MATH 208, Fundamental Mathematics I, connects algebra with quantitative business models. Its Fall 2026 outline moves from functions and financial calculations to matrix methods, resource constraints, counting and probability. The example below develops a production decision from a system of inequalities.

MATH 208 topics to review

Functions, sequences and interest calculationsPresent value, future value and annuitiesLinear systems, elimination and matrix inversesInput–output modelsInequalities and two-variable linear programmingCounting, events and conditional probability

Common difficulties

  • Confusing a resource limit with an equality, then overlooking feasible points where some capacity is unused
  • Choosing a financial or probability formula before identifying the periods, events and assumptions

Preparing for this course

Review fractions, percentages, exponents, equation solving and graphing a line. Practise naming a variable and its units before translating a sentence into mathematics. For financial exercises, distinguish the annual rate from the rate per payment or compounding period. These are study suggestions, not an eligibility assessment.

Registration requirements

The current Concordia calendar lists MATH 206 or equivalent as the prerequisite. Credit and program restrictions also apply; Concordia advising determines equivalencies and eligibility. MATH 209 is linked here as a related course, not a stated prerequisite.

Check the official course catalogue

Try a course problem

Let xx and yy be two unknown real quantities satisfying 3x+2y=263x+2y=26 and x+y=10x+y=10. Find both quantities, then determine whether the same pair also meets the limit 2x+y≤172x+y\le17.

Show the reasoning

1. Eliminate one unknown. Multiply the second equation by 22 to obtain 2x+2y=202x+2y=20. Subtract it from the first equation: (3x+2y)−(2x+2y)=26−20(3x+2y)-(2x+2y)=26-20. The two terms containing yy cancel, leaving x=6x=6. 2. Recover the other unknown. Substitute into x+y=10x+y=10: 6+y=106+y=10, so y=4y=4. The solution pair is (x,y)=(6,4)(x,y)=(6,4). 3. Verify the original equalities. They give 3(6)+2(4)=18+8=263(6)+2(4)=18+8=26 and 6+4=106+4=10, so both hold. 4. Check the additional limit separately. For this pair, 2x+y=2(6)+4=162x+y=2(6)+4=16, and 16≤1716\le17. Thus the pair meets the inequality, with a gap of 17−16=117-16=1 below the limit. If the inequality represents a capacity constraint, that gap is one unit of unused capacity. Solving the equalities alone would not establish whether an additional limit is met.

Use matrices to find a corner, then optimize over the whole feasible region

A resource constraint is a limit, not a promise to use every available hour. This production example connects a small matrix calculation with the graphical method of linear programming.

1. Name the quantities and the objective

A workshop chooses weekly quantities xx of standard batches and yy of premium batches. A standard batch needs 22 assembly hours and 11 finishing hour; a premium batch needs 33 and 22, respectively. There are 120120 assembly hours and 7070 finishing hours available. Assume all produced batches can be sold and each contributes 3030 or 5050 dollars after variable costs. Let Z(x,y)Z(x,y) denote total weekly contribution, in dollars: sales revenue less variable production costs. Each standard batch contributes 30 dollars, so xx batches contribute 30x30x; premium batches similarly contribute 50y50y. Thus contributions add.

2. Translate each resource into a limit

The assembly hours used are 2x+3y2x+3y: multiply each batch count by its hours per batch, then add. This total cannot exceed 120. Finishing similarly uses x+2yx+2y hours, at most 70. Counts cannot be negative. A feasible plan is a pair (x,y)(x,y) satisfying all four inequalities at once. The model is maximize Z=30x+50y\text{maximize }Z=30x+50y subject to 2x+3y≤120,x+2y≤70,x,y≥0.\begin{aligned}2x+3y&\le120,\\x+2y&\le70,\\x,y&\ge0.\end{aligned} Fixed costs do not affect which mix maximizes contribution if they are the same for every mix. We first allow fractional batch quantities; the optimum found below also has whole-number batches.

Feasible region and optimal cornerThe feasible region has corners (0,0), (60,0), (30,20) and (0,35). The maximum of 30x+50y occurs at (30,20).06035xy(30, 20)
Solid: 2x+3y=1202x + 3y = 120; dashed: x+2y=70x + 2y = 70.
The shaded region satisfies both limits and x,y≥0x,y\geq0. Compare the total contribution at all four corners: the intersection is not automatically the best decision.

3. Use a matrix to find the intersection of two boundary lines

A boundary is where one inequality becomes an equality. A corner can occur where two boundaries meet, so first find the meeting point of the two resource lines. In the matrix below, the first and second columns are the coefficients of xx and yy; the column after the bar contains the available hours. Subtracting a multiple of one equation from another preserves their simultaneous solutions. At the intersection where both resources are fully used, solve the boundary equalities. The augmented matrix is [231201270].\left[\begin{array}{cc|c}2&3&120\\1&2&70\end{array}\right]. Replacing row 1 by row 1 minus twice row 2 gives [0−1−201270].\left[\begin{array}{cc|c}0&-1&-20\\1&2&70\end{array}\right]. Thus y=20y=20, then x=30x=30. Check: 2(30)+3(20)=1202(30)+3(20)=120 and 30+2(20)=7030+2(20)=70. This finds one candidate; it does not yet prove that this candidate is best.

4. Find all feasible corners and compare the objective

The nonnegative axes are also boundaries. On the horizontal axis set y=0y=0: the constraints give both x≤60x\le60 and x≤70x\le70, so use the stricter limit 60. On the vertical axis set x=0x=0: both y≤40y\le40 and y≤35y\le35 must hold, so use 35. The feasible polygon therefore has corners (0,0)(0,0), (60,0)(60,0), (30,20)(30,20) and (0,35)(0,35). The other intercepts, (70,0)(70,0) and (0,40)(0,40), violate a constraint. Evaluate Z(0,0)=0,Z(60,0)=1800,Z(30,20)=1900,Z(0,35)=1750.\begin{aligned}Z(0,0)&=0,\\Z(60,0)&=1800,\\Z(30,20)&=1900,\\Z(0,35)&=1750.\end{aligned} This polygon contains feasible points and is bounded: it does not extend indefinitely. A linear objective, such as the total contribution here, attains a maximum at one of its corners, so checking these corners is sufficient. Geometrically, a linear objective has no isolated maximum strictly inside this polygon: along any segment its value changes at a constant rate. Every point in the polygon is a weighted average of corners, so its objective value cannot exceed the largest corner value.

5. Interpret the answer and check its assumptions

Choose 3030 standard batches and 2020 premium batches, for a maximum weekly contribution of 19001900 dollars before fixed costs. Both resource limits are reached in this particular optimum. That conclusion comes from comparing all corners, not from a general rule that every resource must be exhausted. If indivisible batches produced a fractional optimum, simply rounding could break feasibility or miss the best integer choice.

Level 1: Feasibility

Under the original constraints, test (24,24)(24,24) and (25,20)(25,20); compute the contribution only for feasible plans.

Show the explained solution

(24,24)(24,24) uses 120120 assembly hours but 7272 finishing hours, so it is infeasible. The plan (25,20)(25,20) uses 110110 and 6565 hours and contributes 17501750 dollars.

Level 2: New capacity

Change the available hours to 132132 for assembly and 7878 for finishing, keeping Z=30x+50yZ=30x+50y. Find all feasible corners and the best mix.

Show the explained solution

The new boundaries are 2x+3y=1322x+3y=132 and x+2y=78x+2y=78. Subtract twice the second equation from the first: −y=−24-y=-24, hence y=24y=24 and x=78−48=30x=78-48=30. On the axes, both constraints give x≤min⁡(66,78)=66x\le\min(66,78)=66 and y≤min⁡(44,39)=39y\le\min(44,39)=39. The corners are (0,0)(0,0), (66,0)(66,0), (30,24)(30,24), (0,39)(0,39). Their contributions are 00, 19801980, 21002100 and 19501950, respectively; choose (30,24)(30,24).

Level 3: New objective

Return to the original capacities but use Z=40x+50yZ=40x+50y. Find the optimum and any unused capacity.

Show the explained solution

Only the objective changed, so the feasible region and its four corners stay the same. Re-evaluate 40x+50y40x+50y at those corners rather than solving the intersection again. At the original corners, the new contributions are 00, 24002400, 22002200 and 17501750. Choose (60,0)(60,0): assembly uses all 120120 hours, while finishing uses 6060 of 7070, leaving 1010 hours unused. At the optimum, some of a resource can remain unused.

For the next resource-allocation problem, write the inequalities first. Use elimination to locate boundary intersections, reject infeasible points and compare the objective at every remaining corner.

Official Concordia MATH 208 outline, Fall 2026

How tutoring can help

Your tutor can work through your setup to connect matrix rows and columns to the quantities in the problem, then locate an error in elimination or in a resource constraint. For probability, you can compare possible sample spaces together and practise explaining why a counting or conditioning method fits.

Exam preparation for MATH 208

Use your current section outline to choose the scope. Mix finance, systems, inequalities and probability so the method is not supplied by the exercise order. Rework one mistake without its solution, then change one condition and explain how the method or answer changes. Follow your section’s rules for calculators and permitted aids.

We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

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