Difficulty: Intermediate | Prerequisites: Core Concepts doc for Section 2.1, comfort with the vertex formula x = -b/(2a).
These four problems come from textbook pages 136-137 and represent the main types of quadratic modelling you will encounter: maximising area under a perimeter constraint, finding maximum height of a projectile, minimising production cost, and maximising profit. Each follows the same core workflow (constraint, substitution, vertex formula), but the setup differs enough that practising all four builds real fluency. If you can do these without looking at the solutions, you are ready for the unit test.
Four worked problems demonstrating quadratic optimisation: (1) maximise a rectangle's area inside a 200 m running track with semicircular ends, (2) find the peak height of a diver whose path is a downward parabola, (3) find the number of fixtures that minimises a manufacturer's cost, and (4) find the advertising spend that maximises profit. Every solution uses x = -b/(2a).
Problem: An indoor fitness room consists of a rectangular region with a semicircle on each end. The perimeter of the room is a 200-metre single-lane running track. Find the dimensions that maximise the area of the rectangular region.
Let x = the length of each straight side of the rectangle. Let y = the width of the rectangle (which is also the diameter of each semicircle). The radius of each semicircle is therefore r = y/2.
Two semicircles with radius y/2 make one full circle with circumference:
C = 2π(y/2) = πy
The total distance around the track is the two straight sides plus the two semicircular ends:
2x + πy = 200
Solve for y:
y = (200 - 2x) / π
The area of the rectangle is A = x × y. Substitute the expression for y:
A(x) = x × (200 - 2x) / π
A(x) = (200x - 2x²) / π
A(x) = (-2/π)x² + (200/π)x
This is a quadratic in standard form with a = -2/π and b = 200/π.
x = -b/(2a) = -(200/π) / (2 × -2/π) = (200/π) / (4/π) = 200/4 = 50
So x = 50 metres.
y = (200 - 2(50)) / π = 100/π ≈ 31.83 metres.
The rectangular region should be 50 m long and approximately 31.83 m wide. Since a < 0, this is confirmed to be a maximum.
Problem: The path of a diver is given by y = -(4/9)x² + (24/9)x + 12, where y is the height in feet and x is the horizontal distance from the end of the diving board in feet. Find the maximum height.
a = -4/9, b = 24/9, c = 12
Since a < 0, the parabola opens downward, so the vertex is a maximum.
x = -b/(2a) = -(24/9) / (2 × -4/9)
x = -(24/9) / (-8/9)
x = (24/9) × (9/8) = 24/8 = 3
The maximum height occurs at x = 3 feet from the diving board.
y = -(4/9)(3)² + (24/9)(3) + 12
y = -(4/9)(9) + (24/9)(3) + 12
y = -4 + 8 + 12
y = 16 feet
The maximum height of the diver is 16 feet, reached at a horizontal distance of 3 feet from the board. Note that the diver starts at y = 12 feet (the board height when x = 0) and gains 4 feet of elevation before descending.
Problem: A textile manufacturer has daily production costs of C = 100,000 - 110x + 0.045x², where C is the total cost in dollars and x is the number of units produced. How many fixtures should be produced each day to yield a minimum cost?
Rewrite in standard form: C(x) = 0.045x² - 110x + 100,000
a = 0.045, b = -110, c = 100,000
Since a > 0, the parabola opens upward, confirming the vertex is a minimum.
x = -b/(2a) = -(-110) / (2 × 0.045)
x = 110 / 0.09
x ≈ 1,222.2
Since you cannot produce a fraction of a fixture, round to 1,222 fixtures.
Producing approximately 1,222 fixtures per day minimises the daily production cost. The key insight here is that the positive leading coefficient (0.045) guarantees the vertex is a minimum, not a maximum.
Problem: The profit P (in hundreds of dollars) depends on advertising spend x (in hundreds of dollars) according to P = 230 + 20x - 0.5x². What expenditure yields maximum profit?
P(x) = -0.5x² + 20x + 230
a = -0.5, b = 20, c = 230
Since a < 0, the parabola opens downward, so the vertex is a maximum.
x = -b/(2a) = -20 / (2 × -0.5) = -20 / (-1) = 20
Since x is in hundreds of dollars: optimal spend = 20 × $100 = $2,000.
P(20) = -0.5(20)² + 20(20) + 230
P(20) = -0.5(400) + 400 + 230
P(20) = -200 + 400 + 230 = 430
Since P is in hundreds of dollars: maximum profit = 430 × $100 = $43,000.
Spending $2,000 on advertising yields the maximum profit of $43,000. Pay attention to units here: both x and P are measured in hundreds of dollars, so multiply by 100 when stating the final answer.
In Problem 76, students sometimes use the full circumference formula (2πr) for each semicircle, doubling the curved distance. Two semicircles together make one full circle, so the combined curved distance is πy (one circumference with diameter y).
In Problem 77, students sometimes confuse the starting height (c = 12) with the maximum height. The vertex gives the maximum, which is 16 feet, not 12.
In Problem 80, the equation as given starts with 100,000 - 110x + 0.045x². Students sometimes misread the sign of b. Rewriting in standard form (0.045x² - 110x + 100,000) clarifies that b = -110.
In Problem 82, the most common error is forgetting that x and P are in hundreds of dollars. The vertex gives x = 20, but the answer is $2,000 in advertising, and the maximum profit is $43,000, not $430.
Q: A projectile's height is modelled by h(t) = -5t² + 30t + 2. At what time does it reach its maximum height, and what is that height?
A: t = -30/(2 × -5) = 3 seconds. h(3) = -5(9) + 30(3) + 2 = -45 + 90 + 2 = 47 metres.
Q: A company's revenue is R(x) = -3x² + 48x, where x is the price in dollars. What price maximises revenue?
A: x = -48/(2 × -3) = 8 dollars. Revenue at that price: R(8) = -3(64) + 48(8) = -192 + 384 = $192.
Q: You have 60 metres of fencing to enclose a rectangular garden against a wall (three sides). What dimensions maximise area?
A: Constraint: 2w + l = 60, so l = 60 - 2w. Area = w(60 - 2w) = -2w² + 60w. w = -60/(2 × -2) = 15 m, l = 30 m. Max area = 450 m².
Q: In Problem 76, why is the answer x = 50 a maximum rather than a minimum?
A: Because a = -2/π, which is negative. A negative leading coefficient means the parabola opens downward, so the vertex is the highest point on the curve, which corresponds to the greatest area.
quadratic word problems worked examples, honors precalculus 2.1, maximize area semicircle rectangle, diver path maximum height, minimize production cost quadratic, maximize profit advertising quadratic, vertex formula examples, x = -b/2a worked problems, constraint equation examples, perimeter optimization, projectile motion quadratic, textbook pages 136-137 solutions