Represent constraints from more complicated situations with equations, inequalities and systems that mix linear, quadratic and exponential relationships, and decide which solutions are viable

Learning goal: Represent the constraints of a situation by equations or inequalities, and by systems of equations and/or inequalities, in problems that draw from more complicated situations than a single linear relationship, where linear, quadratic and exponential functions appear together; and interpret each solution as a viable or non-viable option in the modeling context.

Constraints, systems and viable options. A constraint is a condition that the situation places on the variables. A condition that says "exactly" is written as an equation; conditions that say "at most", "at least", "no more than" or "cannot exceed" are written as inequalities. When several constraints must hold at the same time they form a system, and a solution of the system is a choice of values that satisfies every one of them. In a modeling context a solution of the algebra is still only a candidate. It is a viable option if it also makes sense in the situation: a whole number where objects are counted, a positive length, a time after the start. It is non-viable if it breaks any constraint, stated or unstated.

The linear starting point. A student builds a snack from x servings of granola and y servings of yogurt. With hypothetical figures of $0.50 and 3 g of protein per serving of granola and $0.80 and 6 g of protein per serving of yogurt, a budget of $4.00 and a target of at least 18 g of protein give the system 0.50x + 0.80y ≤ 4.00, 3x + 6y ≥ 18, x ≥ 0, y ≥ 0. The combination (2, 3) costs $3.40 and supplies 24 g, so it is viable. The combination (1, 2) costs only $2.10 but supplies 15 g, so it is non-viable; (6, 2) supplies 30 g but costs $4.60, so it is non-viable too. Every constraint here is linear: each side changes by a constant amount per serving.

What makes a situation more complicated. In many real problems the functions involved are still linear, quadratic or exponential, but the situation combines them. A constraint may be quadratic (an area, a revenue, a height) or exponential (a population, a balance, a decaying amount). A system may mix types, so that a line has to be compared with a parabola or with an exponential curve. Some constraints are never stated because they are obvious to a person and invisible to the algebra: lengths are positive, people are counted in whole numbers, time starts at zero. The algebra may then produce solutions that are not options at all. And some mixed systems cannot be solved by rearranging; a table or a graph is the method.

A model worked in full. A hypothetical community garden is to be a rectangle x meters wide and y meters long. The group has 60 m of fencing and wants an area of at least 200 m². The plot of land is only 14 m wide but more than long enough for any length considered here, and x is the side that runs across the plot's width. Write every constraint. Fencing: 2x + 2y ≤ 60, which simplifies to x + y ≤ 30, a linear inequality. Area: xy ≥ 200, which is not linear, because the variables are multiplied. Width: x ≤ 14. Unstated: x > 0 and y > 0. These five inequalities are the system.

Test some candidates. (12, 18): 12 + 18 = 30 ≤ 30, area 216 ≥ 200, width 12 ≤ 14, so it is viable. (8, 22): the fencing is fine, but the area is 176, so it is non-viable. (16, 14): the fencing is fine and the area is 224, but 16 m is wider than the plot, so it is non-viable. (13, 16): 29 ≤ 30 and area 208, so it is viable with 2 m of fencing left over.

To find all the viable widths when the whole 60 m is used, substitute the equation y = 30 − x into the area constraint: x(30 − x) ≥ 200. Expanding and collecting gives −x² + 30x − 200 ≥ 0, and multiplying by −1 reverses the inequality: x² − 30x + 200 ≤ 0, or (x − 10)(x − 20) ≤ 0. A product is negative or zero exactly when one factor is zero or the factors have opposite signs, which happens for 10 ≤ x ≤ 20. Test values confirm it: x = 5 gives (−5)(−15) = 75, positive; x = 15 gives (5)(−5) = −25, negative; x = 25 gives (15)(5) = 75, positive. So the quadratic constraint allows widths from 10 m to 20 m, and the linear constraint x ≤ 14 cuts that down to 10 ≤ x ≤ 14. The largest area the algebra offers is at x = 15, where x(30 − x) = 225 m², but x = 15 is non-viable. On the interval that is viable the area increases with x, so the best viable option is at the boundary: 14 m by 16 m, with area 224 m².

Exponential constraints. When the variable is in an exponent and stands alone, logarithms solve the constraint. A hypothetical pond is stocked with 200 fish and the population grows 12% a year, so F = 200(1.12)ᵗ. The owner will open the pond for fishing in a year when it holds at least 400 fish and no more than the 600 it can support. The constraints 400 ≤ 200(1.12)ᵗ ≤ 600 become 2 ≤ 1.12ᵗ ≤ 3. Taking logarithms and dividing by ln 1.12, which is positive because 1.12 is greater than 1 (so the inequality signs keep their direction), gives ln 2 ÷ ln 1.12 ≤ t ≤ ln 3 ÷ ln 1.12, that is 6.12 ≤ t ≤ 9.69. If the fish are counted once a year, t is a whole number and the viable opening years are 7, 8 and 9 (442, 495 and 555 fish). Year 6, with 395 fish, is non-viable although 6 is the nearest whole number to 6.12: the constraint decides the direction of rounding, not the usual rounding rule. When an exponential expression and a linear expression both contain the variable, as in the second worked example below, no rearrangement with the algebra of this course isolates the variable, and a table or a graph is used instead.

The misconceptions. The first is to treat every solution of the algebra as an answer to the problem. The garden's largest area, a negative time, a break-even point beyond a workshop's capacity: each satisfies some equation and none is an option. Always return to the situation and test the candidate against every constraint. The second is to solve a quadratic inequality as if it were linear, writing x² − 30x + 200 ≤ 0 as "x ≤ 10 or x ≤ 20". A quadratic inequality is solved from the signs of its factors or from its graph: the parabola y = x² − 30x + 200 opens upward and is on or below the axis only from one zero to the other, so the solution is the interval from 10 to 20, not everything to the left of them.

Worked linear and quadratic system: A hypothetical workshop makes x lamps a month. To sell x lamps it must price them at 120 − 2x dollars each, so its revenue is R = x(120 − 2x), a quadratic function. Its costs are C = 800 + 20x dollars, a linear function. It cannot make more than 35 lamps a month, the owner will not price a lamp below $60, and x is a whole number. Break-even points solve the system R = x(120 − 2x), R = 800 + 20x: 120x − 2x² = 800 + 20x gives 2x² − 100x + 800 = 0, so x² − 50x + 400 = 0, (x − 10)(x − 40) = 0, and x = 10 or x = 40. Check: at x = 10 revenue and cost are both $1,000; at x = 40 both are $1,600. The workshop avoids a loss when R ≥ C, which is x² − 50x + 400 ≤ 0, so 10 ≤ x ≤ 40. Now the other constraints: capacity gives x ≤ 35, and the price floor 120 − 2x ≥ 60 gives x ≤ 30. The viable options are the whole numbers from 10 to 30. The solution x = 40 is non-viable twice over: it exceeds capacity, and it needs a price of $40. Profit is R − C = −2x² + 100x − 800, with vertex at x = 100 ÷ 4 = 25, where the profit is −1,250 + 2,500 − 800 = 450 dollars at a price of $70; 25 is viable, so it is the best option. If the price floor were $80 instead, the constraint 120 − 2x ≥ 80 would give x ≤ 20, the vertex would be non-viable, and the best viable option would be x = 20, with profit −800 + 2,000 − 800 = 400 dollars.

Worked exponential and linear system: A hypothetical city of 40,000 people grows 3% a year, so its population after t years is P = 40,000(1.03)ᵗ. Its water plant can serve 46,000 people now and is being expanded to serve 1,500 more each year, so capacity is W = 46,000 + 1,500t. The constraint is P ≤ W with t ≥ 0. Both sides contain t and one of them has t in an exponent, so use a table. At t = 0: 40,000 against 46,000. At t = 10: 53,757 against 61,000. At t = 20: 72,244 against 76,000. At t = 24: 81,312 against 82,000. At t = 25: 83,751 against 83,500. The constraint holds through year 24 and fails in year 25. After that it stays broken: the population is then growing by about 2,500 people a year (3% of 83,751), more than the 1,500 added to capacity, and that gap keeps widening. A graph of the two functions shows that the equation P = W has two solutions, near t = 24.7 and near t = −11.9. The negative solution is non-viable, because the model starts at t = 0. The viable years are the whole numbers 0 through 24, and the practical conclusion is that the expansion must speed up before year 25.

Worked system in two variables with a quadratic boundary: A hypothetical road tunnel has a parabolic arch: at a horizontal distance x meters from the center line the roof is 5 − 0.2x² meters above the road. A vehicle w meters wide and h meters tall drives along the center line, so its top corners are at x = w ÷ 2, and it must keep 0.5 m of vertical clearance between those corners and the roof: h + 0.5 ≤ 5 − 0.2(w ÷ 2)², which simplifies to h ≤ 4.5 − 0.05w². Hypothetical road rules add two linear constraints, w ≤ 2.6 and h ≤ 4.2, and both measurements are positive. Test three vehicles. A truck 2.5 m wide and 4.0 m tall: the arch allows 4.5 − 0.05(6.25) = 4.1875 m, so 4.0 fits and both rules are met; viable. A truck 2.6 m wide and 4.2 m tall meets both rules, but the arch allows only 4.5 − 0.05(6.76) = 4.162 m; non-viable. A load 4.0 m wide and 3.6 m tall fits the arch, which allows 4.5 − 0.8 = 3.7 m, but breaks the width rule; non-viable. Which constraint limits the height depends on the width: 4.5 − 0.05w² ≥ 4.2 exactly when w² ≤ 6, that is w ≤ 2.45 (to two decimals), so for narrower vehicles the 4.2 m rule is the tighter constraint and for vehicles between 2.45 m and 2.6 m wide the arch is.

Represent the constraints: For the garden, draw axes with the width x from 0 to 30 on the horizontal axis and the area on the vertical axis. Graph A = x(30 − x) for the case in which all 60 m of fencing is used, draw the horizontal line A = 200, and mark the points where they meet. Shade the part of the x-axis where the area constraint holds, then mark x = 14 and shade, in a second color, the widths that satisfy every constraint. Under the graph, list the five inequalities of the system and label each one linear or not linear.

Independent performance: Write and solve your own constraint problem from a more complicated situation. It must include at least one linear constraint, at least one quadratic or exponential constraint, and at least one unstated constraint that you state (whole numbers, positive values or a starting time). Suitable situations include an enclosure with limited fencing and a minimum area, a product whose revenue is quadratic and whose cost is linear, or a growing quantity that must stay between two limits. (1) Define each variable with its unit. (2) Write the system and say which constraints are equations and which are inequalities. (3) Solve it by algebra where that is possible and by a table or graph where it is not, and say which method you used and why. (4) Give two viable options and two non-viable options, naming for each non-viable option the constraint it breaks. (5) State the best viable option for a goal you choose, and say whether it lies at a vertex or on a boundary. Success means every constraint in the story appears in the system, each solution is tested against all constraints rather than one, non-viable solutions of the algebra are identified and explained, and the final answer is a sentence with units.