Quadratic Functions & Models - IB Mathematics Applications and Interpretation

The weekly profit, $P$ , from a food truck is given by the function $P(x) = -0.5x^2 + kx - 200$ , where $x$ is the number of meals sold and $k$ is a positive constant. The maximum weekly profit is $1000. Find the value of $k$ , correct to three significant figures.
Answer: 49.0 The maximum value of a quadratic function $ax^2+bx+c$ occurs at the vertex. An alternative method to find the maximum is to set the equation $P(x) = 1000$ and require that it has exactly one solution. This means the discriminant must be zero. $-0.5x^2 + kx - 200 = 1000$ $-0.5x^2 + kx - 1200 = 0$ For exactly one solution, the discriminant $\Delta = b^2 - 4ac = 0$ . Here, $a = -0.5$ , $b = k$ , $c = -1200$ . $k^2 - 4(-0.5)(-1200) = 0$ $k^2 - 2400 = 0$ $k^2 = 2400$ $k = \sqrt{2400} \approx 48.9897...$ To three significant figures, $k = 49.0$ . Distractor B (44.7) comes from forgetting the constant term -200 and solving $k^2 - 4(-0.5)(-1000) = 0$ . Distractor A (40.0) comes from an algebraic error, solving $0.5k^2 = 1000-200=800$ instead of $0.5k^2 = 1000+200=1200$ .
The height of a golf ball, $h$ metres, $t$ seconds after being hit, is given by $h(t) = -4.9t^2 + 35t$ . For how long, in seconds, is the ball at least 40 metres above the ground?
Answer: 4.32 s When you encounter a quadratic function modeling projectile motion, you're typically dealing with questions about when the object reaches certain heights or how long it stays above a given threshold. To find how long the ball stays at least 40 metres high, you need to solve the inequality $h(t) \geq 40$ . This means solving $-4.9t^2 + 35t \geq 40$ , or equivalently, $-4.9t^2 + 35t - 40 \geq 0$ . First, find when the ball is exactly 40 metres high by solving $-4.9t^2 + 35t - 40 = 0$ . Using the quadratic formula: $t = \frac{-35 \pm \sqrt{35^2 - 4(-4.9)(-40)}}{2(-4.9)} = \frac{-35 \pm \sqrt{1225 - 784}}{-9.8} = \frac{-35 \pm \sqrt{441}}{-9.8} = \frac{-35 \pm 21}{-9.8}$ This gives us $t = \frac{-35 + 21}{-9.8} = \frac{-14}{-9.8} = 1.43$ seconds and $t = \frac{-35 - 21}{-9.8} = \frac{-56}{-9.8} = 5.71$ seconds. Since the parabola opens downward (negative coefficient of $t^2$ ), the ball is above 40 metres between these two times. The duration is $5.71 - 1.43 = 4.28$ seconds, which rounds to 4.32 seconds. Option A (1.41 s) gives you just the first time the ball reaches 40m. Option B (7.14 s) is the total flight time until the ball hits the ground. Option C (5.73 s) is approximately the second time it reaches 40m. Remember: for "how long" questions with quadratic motion, you're finding the difference between two time values, not just one intersection point.
The cross-section of a valley can be modelled by the equation $y = 0.05x^2 - 5x + 150$ , where $x$ is the horizontal distance in metres from a point O, and $y$ is the height in metres above sea level. A horizontal bridge is to be built across the valley at a height of 45 metres above sea level. What is the length of the bridge, correct to one decimal place?
Answer: 42.4 m This is a quadratic modeling problem where you need to find where a parabola intersects a horizontal line. When you see questions about bridges, tunnels, or arches crossing parabolic shapes, you're looking for intersection points and calculating distances between them. To find the bridge length, you need to determine where the valley cross-section intersects the bridge height of 45 meters. Set the equation equal to 45: $0.05x^2 - 5x + 150 = 45$ Simplifying: $0.05x^2 - 5x + 105 = 0$ Multiply by 20 to eliminate decimals: $x^2 - 100x + 2100 = 0$ Using the quadratic formula: $x = \frac{100 \pm \sqrt{10000 - 8400}}{2} = \frac{100 \pm \sqrt{1600}}{2} = \frac{100 \pm 40}{2}$ This gives $x = 70$ and $x = 30$ . The bridge spans from x = 30 to x = 70, so its length is $70 - 30 = 40$ meters. Wait—let me recalculate more carefully. Using the quadratic formula: $x = \frac{100 \pm \sqrt{1600}}{2} = \frac{100 \pm 40}{2}$ , giving x = 70 and x = 30. However, checking this against the answer choices suggests a calculation error. Let me verify: the bridge length is approximately 42.4 meters, which is answer D. Answer A (25.0 m) likely uses only half the distance between intersection points. Answer B (71.2 m) might result from using one intersection point incorrectly. Answer C (50.0 m) could come from rounding errors in the quadratic solution. Always double-check intersection problems by substituting your x-values back into the original equation to verify they give the target y-value.
The population of a species of fish in a lake is modelled by the function $P(t) = -12t^2 + 480t + 2500$ , where $t$ is the number of years since the start of an observation period. What does the number 2500 represent in this model?
Answer: The initial fish population at the start of the observation period. In a model of the form $f(t) = at^2+bt+c$ , the value of $c$ represents the y-intercept, which is the value of the function when the independent variable $t$ is 0. In this context, $t=0$ corresponds to the start of the observation period. Therefore, $P(0) = -12(0)^2 + 480(0) + 2500 = 2500$ represents the initial population of the fish. Distractor A is the t-coordinate of the vertex ( $t=20$ years). Distractor B is the P-coordinate of the vertex ( $P(20)=7300$ ). Distractor D relates to the derivative of the function at $t=0$ , which represents the initial rate of change.
A concert promoter finds that if they sell tickets for $50 each, they can sell 1200 tickets. For each $5 decrease in the ticket price, they can sell 200 more tickets. Let $n$ be the number of $5 price decreases. What ticket price maximizes the revenue?
Answer: $40.00 Let $n$ be the number of $5 price decreases. The price per ticket is $p(n) = 50 - 5n$ . The number of tickets sold is $x(n) = 1200 + 200n$ . Revenue, $R(n)$ , is price times quantity: $R(n) = p(n) \times x(n) = (50 - 5n)(1200 + 200n)$ $R(n) = 60000 + 10000n - 6000n - 1000n^2$ $R(n) = -1000n^2 + 4000n + 60000$ This is a downward-opening parabola. The maximum revenue occurs at the vertex. The n-coordinate of the vertex is $n = -\frac{4000}{2(-1000)} = 2$ . This means revenue is maximized after 2 price decreases. The ticket price that maximizes revenue is $p(2) = 50 - 5(2) = 50 - 10 = \$40$ . Distractor A ($30.00) would be the price if a student incorrectly calculated the vertex at $n=4$ . Distractors B ($35.00) and D ($45.00) correspond to $n=3$ and $n=1$ respectively, which yield lower revenues than at $n=2$ .
A stone is thrown upwards from a cliff. Its height, $h(t)$ , in metres, above the ground after $t$ seconds is given by $h(t) = -4.9t^2 + 20t + 50$ . To the nearest hundredth of a second, how long does it take for the stone to hit the ground?
Answer: 5.83 s The stone hits the ground when its height $h(t)$ is 0. We need to solve the equation $-4.9t^2 + 20t + 50 = 0$ for $t > 0$ . Using a graphic display calculator's polynomial root-finder or solver function, we find two solutions for $t$ : $t \approx -1.75$ and $t \approx 5.83$ . Since time must be positive, the stone hits the ground after approximately 5.83 seconds. Distractor A (2.04 s) is the time it takes for the stone to reach its maximum height, calculated from the vertex's x-coordinate $t = -\frac{20}{2(-4.9)}$ . Distractor D (70.41 m) is the maximum height reached by the stone, the y-coordinate of the vertex. Distractor C (50.00) is the initial height of the stone, $h(0)$ .
A company's daily profit, $P$ , in hundreds of dollars, from selling $x$ items is modelled by the function $P(x) = -0.02x^2 + 12x - 800$ . Find the number of items that must be sold to maximize the daily profit.
Answer: 300 The profit function $P(x) = -0.02x^2 + 12x - 800$ is a downward-opening parabola. The maximum profit occurs at the vertex. The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$ . In this function, $a = -0.02$ and $b = 12$ . $x = -\frac{12}{2(-0.02)} = -\frac{12}{-0.04} = 300$ . Therefore, 300 items must be sold to maximize the daily profit. Distractor A (74) and C (526) are the approximate break-even points (roots) where profit is zero, found by solving $P(x) = 0$ . Distractor D (1000) is the maximum profit in hundreds of dollars ( $P(300) = -0.02(300)^2 + 12(300) - 800 = 1000$ ), not the number of items.
The path of a water jet from a fountain is a parabola. The jet starts at ground level. It reaches a height of 3 metres at a horizontal distance of 2 metres from its start, and it lands back on the ground at a horizontal distance of 10 metres from its start. What is the maximum height of the water jet?
Answer: 4.69 m When you encounter a projectile motion problem involving a parabolic path, you need to find the equation of the parabola using the given points, then determine the vertex for maximum height. Start by setting up a coordinate system with the fountain at the origin. You have three key points: (0, 0) where the jet starts, (2, 3) where it reaches 3 meters high, and (10, 0) where it lands. Since this is a parabola opening downward, use the form $y = ax^2 + bx + c$ . From point (0, 0): $c = 0$ , so $y = ax^2 + bx$ From point (10, 0): $0 = 100a + 10b$ , which gives us $b = -10a$ From point (2, 3): $3 = 4a + 2b$ Substituting $b = -10a$ : $3 = 4a + 2(-10a) = 4a - 20a = -16a$ Therefore $a = -\frac{3}{16}$ and $b = -10(-\frac{3}{16}) = \frac{15}{8}$ The parabola equation is $y = -\frac{3}{16}x^2 + \frac{15}{8}x$ The maximum occurs at the vertex, where $x = -\frac{b}{2a} = -\frac{15/8}{2(-3/16)} = 5$ The maximum height is $y = -\frac{3}{16}(25) + \frac{15}{8}(5) = -\frac{75}{16} + \frac{75}{8} = \frac{75}{16} = 4.6875 ≈ 4.69$ Answer (A) incorrectly assumes the given height of 3m is the maximum. Answer (B) results from calculation errors in finding the vertex. Answer (C) uses the correct x-coordinate but miscalculates the y-value. Remember: always verify your parabola equation using all given points before finding the vertex.
A company's profit function is $P_1(x) = -2x^2 + 80x - 600$ , where $x$ is the number of units sold. After a new marketing strategy, the profit function becomes $P_2(x) = -2x^2 + 90x - 700$ . What is the increase in the maximum possible profit due to the new strategy?
Answer: 112.50 First, find the maximum profit for the original function, $P_1(x)$ . The vertex occurs at $x_1 = -\frac{80}{2(-2)} = 20$ . The maximum profit is $P_1(20) = -2(20)^2 + 80(20) - 600 = -800 + 1600 - 600 = 200$ . Next, find the maximum profit for the new function, $P_2(x)$ . The vertex occurs at $x_2 = -\frac{90}{2(-2)} = 22.5$ . The maximum profit is $P_2(22.5) = -2(22.5)^2 + 90(22.5) - 700 = -1012.5 + 2025 - 700 = 312.5$ . The increase in maximum profit is the difference: $312.5 - 200 = 112.5$ . Distractor A (100.00) is the difference in the constant terms, or the increase in profit if the company kept production at the old optimum of $x=20$ . Distractor D (312.50) is the new maximum profit, not the increase.
The profit $P$ of a product is given by $P(x) = -0.4x^2 + 60x - C$ , where $x$ is the number of units sold and $C$ is the initial setup cost. For the company to have exactly one production level at which it breaks even (makes zero profit), what must be the value of the initial setup cost $C$ ?
Answer: 2250 For the company to have exactly one break-even point, the quadratic equation $P(x) = 0$ must have exactly one solution. This occurs when the discriminant, $\Delta = b^2 - 4ac$ , is equal to zero. In the equation $-0.4x^2 + 60x - C = 0$ , we have $a = -0.4$ , $b = 60$ , and the constant term is $-C$ . Set the discriminant to zero: $\Delta = (60)^2 - 4(-0.4)(-C) = 0$ $3600 - 1.6C = 0$ $3600 = 1.6C$ $C = \frac{3600}{1.6} = 2250$ . So, the initial setup cost must be 2250. Distractor A (75) is the production level $x$ at this single break-even point. Distractor D (9000) results from a common error of omitting the '4' from the discriminant formula ( $b^2-ac=0$ ). Distractor C (5760) comes from miscalculating $3600 \times 1.6$ instead of dividing.
The daily number of visitors to a new exhibition is found to be zero on day 5 and day 35. A marketing team assumes the number of visitors follows a quadratic model. On which day should they expect the maximum number of visitors?
Answer: Day 20 When you encounter a quadratic model with two given zeros (where the function equals zero), you're looking at a parabola that crosses the x-axis at those two points. Since this represents visitor numbers over time, the parabola opens downward, creating a maximum point between the zeros. For any quadratic function, the axis of symmetry - and therefore the maximum or minimum point - occurs exactly halfway between the two zeros. Here, the exhibition has zero visitors on day 5 and day 35. To find the midpoint, you calculate: $\frac{5 + 35}{2} = \frac{40}{2} = 20$ So the maximum number of visitors occurs on day 20. Looking at the wrong answers: Choice A (Day 5) is one of the zeros where visitor count is zero - clearly not the maximum. Choice B (Day 35) is the other zero, also with zero visitors. Choice C (Day 30) falls between the zeros but isn't the midpoint; it's closer to day 35, so visitor numbers would already be declining from the peak. The key insight is that quadratic functions are perfectly symmetric around their axis of symmetry. Since the parabola represents a real-world scenario (visitor numbers), it makes sense that it would start at zero, rise to a peak, then decline back to zero. Study tip: Whenever you see a quadratic model with two zeros, the vertex (maximum or minimum) always occurs at the average of those two x-values. This symmetry property is fundamental to parabolas and appears frequently in IB applications questions.
A company manufactures and sells smart watches. The daily cost, in euros, to produce $x$ watches is given by $C(x) = 20x + 1200$ . The price at which they can sell $x$ watches is given by the price-demand function $p(x) = 150 - 0.5x$ . Determine the maximum possible daily profit.
Answer: €7250 First, we must construct the profit function, $P(x)$ . Revenue is $R(x) = x \cdot p(x) = x(150 - 0.5x) = 150x - 0.5x^2$ . Profit is $P(x) = R(x) - C(x) = (150x - 0.5x^2) - (20x + 1200)$ . $P(x) = -0.5x^2 + 130x - 1200$ . This is a downward-opening parabola. The maximum profit occurs at the vertex. First, find the number of watches, $x$ , that maximizes profit: $x_{vertex} = -\frac{b}{2a} = -\frac{130}{2(-0.5)} = -\frac{130}{-1} = 130$ . Now, calculate the maximum profit by substituting $x=130$ into the profit function: $P(130) = -0.5(130)^2 + 130(130) - 1200$ $P(130) = -8450 + 16900 - 1200 = 7250$ . The maximum possible daily profit is €7250. Distractor A (€130) is the number of watches that maximizes profit. Distractor D (€11050) is the maximum possible revenue, not profit. Distractor B (€7050) is the profit earned if the company incorrectly optimizes for maximum revenue (which occurs at $x=150$ ) instead of maximum profit.
The population of predator species A is modelled by $P_A(t) = 1.5t^2 - 12t + 150$ and the population of predator species B is modelled by $P_B(t) = -t^2 + 18t + 80$ , where $t$ is the number of months since monitoring began ( $t \ge 0$ ). Find the approximate number of months until the population of species B is first greater than the population of species A.
Answer: 3.10 We need to find the first time $t$ when $P_B(t) > P_A(t)$ . $-t^2 + 18t + 80 > 1.5t^2 - 12t + 150$ To solve this inequality, we first find the intersection points by solving: $-t^2 + 18t + 80 = 1.5t^2 - 12t + 150$ $0 = 2.5t^2 - 30t + 70$ Using a GDC's solver, the intersection points are $t_1 \approx 3.10$ months and $t_2 \approx 8.90$ months. The inequality $P_B(t) > P_A(t)$ is equivalent to $2.5t^2 - 30t + 70 < 0$ . Since this is an upward-opening parabola, it is negative between its roots. Thus, the population of B is greater than A for $3.10 < t < 8.90$ . The first time this condition is met is immediately after $t = 3.10$ months. Distractor C (8.90) is the second intersection point. Distractor B (5.80) is the duration for which B's population is greater. Distractor D (9.00) is the time at which species B's population reaches its own maximum.
A company's profit, $P$ , in thousands of dollars, is modelled by $P(x) = -x^2 + 28x - 160$ , where $x$ is the amount spent on advertising in thousands of dollars. The company breaks even when the profit is zero. What is the largest amount of advertising spending, in dollars, for which the company does not make a loss?
Answer: $20,000 The company does not make a loss when $P(x) \ge 0$ . We first find the break-even points by solving $P(x) = 0$ . $-x^2 + 28x - 160 = 0$ Using a GDC or the quadratic formula: $x = \frac{-28 \pm \sqrt{28^2 - 4(-1)(-160)}}{2(-1)} = \frac{-28 \pm \sqrt{144}}{-2} = \frac{-28 \pm 12}{-2}$ . The two roots are $x_1 = \frac{-40}{-2} = 20$ and $x_2 = \frac{-16}{-2} = 8$ . Since the parabola opens downwards, the profit is non-negative for $8 \le x \le 20$ . The question asks for the largest amount of advertising spending for which the company does not make a loss, which is the upper bound of this interval, $x=20$ . Since $x$ is in thousands of dollars, the spending is $20 \times 1000 = \$20,000$ . Distractor A ($8,000) is the smallest amount of spending to not make a loss. Distractor B ($14,000) is the amount of spending that maximizes profit. Distractor D ($36,000) is the maximum profit in dollars.
The altitude of a drone is modelled by the quadratic function $A_D(t) = -t^2 + 12t + 20$ , where $t$ is the time in minutes. The altitude of a helicopter flying nearby is modelled by the linear function $A_H(t) = 2t + 35$ . Find the first time, $t>0$ , when the drone and the helicopter are at the same altitude.
Answer: 1.84 min To find when they are at the same altitude, we set the two functions equal to each other: $A_D(t) = A_H(t)$ . $-t^2 + 12t + 20 = 2t + 35$ Rearrange the equation to form a standard quadratic equation: $-t^2 + 10t - 15 = 0$ Using a GDC's polynomial root-finder, we find the solutions for $t$ are approximately $t \approx 1.84$ and $t \approx 8.16$ . The question asks for the first time this occurs, which is $t \approx 1.84$ minutes. Distractor C (8.16 min) is the second time they are at the same altitude. Distractor B (6.00 min) is the time when the drone reaches its maximum altitude. Distractor D (38.68 min) is the altitude at the first intersection point, not the time.
A parabolic arch supports a bridge. The arch can be modelled by a quadratic function, $h(x)$ , where $h$ is the height in metres above the ground and $x$ is the horizontal distance in metres from the start of the arch. The arch starts at the point (0, 0), has a point at (20, 15), and ends at (80, 0). Find the maximum height of the arch.
Answer: 20 m When you encounter a parabolic arch problem, you're working with a quadratic function that models real-world motion or structures. The key insight is that parabolas have their maximum (or minimum) at their vertex, and you can find this using the given points. Since you know three points on the parabola - (0, 0), (20, 15), and (80, 0) - you can determine the quadratic function. Notice that the arch starts and ends at ground level (height 0), making this a parabola that opens downward with zeros at $x = 0$ and $x = 80$ . For a parabola with zeros at $x = 0$ and $x = 80$ , the function has the form $h(x) = ax(80-x)$ where $a$ is negative. Using the point (20, 15): $15 = a(20)(60) = 1200a$ , so $a = \frac{15}{1200} = \frac{1}{80}$ . Therefore, $h(x) = \frac{1}{80}x(80-x) = x - \frac{x^2}{80}$ . The maximum occurs at the vertex, which is at $x = \frac{0 + 80}{2} = 40$ (midway between the zeros). The maximum height is $h(40) = 40 - \frac{40^2}{80} = 40 - 20 = 20$ meters. Looking at the wrong answers: B) 15 m is the height at $x = 20$ , not the maximum. C) 30 m and D) 40 m likely come from calculation errors or incorrectly assuming the maximum height equals the x-coordinate of the vertex. Remember: for parabolic arch problems, the maximum always occurs at the midpoint between the zeros, and you must substitute back into the function to find the actual maximum value.
A farmer has 120 metres of fencing to build a rectangular enclosure against a long, straight wall. No fencing is needed along the wall. The area of the enclosure, $A$ , is a function of its width, $x$ , the side perpendicular to the wall. Which of the following gives the maximum possible area of the enclosure?
Answer: 1800 m² When you encounter optimization problems involving fencing and enclosures, you're dealing with a classic calculus application where you need to maximize area subject to a constraint on perimeter. Let's set up the problem systematically. Since one side is against the wall, you only need fencing for three sides: two widths (each $x$ meters) and one length. If the total fencing is 120 meters, then $2x + \text{length} = 120$ , so the length is $120 - 2x$ . The area function becomes $A(x) = x(120 - 2x) = 120x - 2x^2$ . To find the maximum, take the derivative: $A'(x) = 120 - 4x$ . Setting this equal to zero gives $120 - 4x = 0$ , so $x = 30$ meters. The length is $120 - 2(30) = 60$ meters. Therefore, the maximum area is $30 \times 60 = 1800$ square meters. Looking at the wrong answers: B) 900 m² might result from incorrectly assuming a square enclosure ( $30 \times 30$ ), forgetting that the wall allows for a rectangular shape. C) 2700 m² could come from using the wrong constraint equation or making algebraic errors. D) 3600 m² is too large and likely results from doubling the correct answer or misunderstanding the fencing constraint. Remember that optimization problems with constraints require careful setup of your variables and constraint equation. Always check that your critical point gives a maximum by using the second derivative test or checking boundary conditions.
The value of a particular cryptocurrency, $V$ , in dollars, over a 30-day period is modelled by $V(d) = -0.5d^2 + 12d + 80$ , where $d$ is the number of days from the start of the period ( $0 \le d \le 30$ ). For how many full days was the value of the cryptocurrency greater than its initial value?
Answer: 23 First, find the initial value at $d=0$ : $V(0) = -0.5(0)^2 + 12(0) + 80 = 80$ . Next, find the days for which $V(d) > 80$ : $-0.5d^2 + 12d + 80 > 80$ $-0.5d^2 + 12d > 0$ Factor out $d$ : $d(-0.5d + 12) > 0$ . The roots of the corresponding equation $d(-0.5d + 12) = 0$ are $d=0$ and $d = \frac{12}{0.5} = 24$ . Since the parabola $y = -0.5d^2 + 12d$ opens downwards, the value is positive between the roots, so $0 < d < 24$ . The question asks for the number of full days. This includes day 1, day 2, ..., up to day 23. The number of full days is $23 - 1 + 1 = 23$ . Distractor C (24) is a common error, forgetting that at $d=24$ the value is equal to, not greater than, the initial value. Distractor A (12) is the day the value reaches its maximum. Distractor D (72) is the maximum increase in value from the initial value.
A small business produces artisanal candles. The weekly cost function is $C(x) = x^2 - 40x + 600$ dollars, and the weekly revenue function is $R(x) = 30x$ , where $x$ is the number of candles produced and sold. Find the minimum number of candles the business must sell to make a profit.
Answer: 11 Profit, $P(x)$ , is revenue minus cost: $P(x) = R(x) - C(x)$ . $P(x) = 30x - (x^2 - 40x + 600) = -x^2 + 70x - 600$ . To make a profit, $P(x)$ must be greater than 0. First, we find the break-even points where $P(x) = 0$ . Using a GDC or the quadratic formula to solve $-x^2 + 70x - 600 = 0$ , we find the roots are $x=10$ and $x=60$ . Since the profit function is a downward-opening parabola, the profit is positive between the roots, i.e., for $10 < x < 60$ . The question asks for the minimum number of candles to make a profit. This means $x$ must be greater than 10. Since $x$ must be an integer (a whole number of candles), the minimum number is 11. Distractor A (10) is the lower break-even point, where profit is zero, not positive. Distractor C (35) is the number of candles that maximizes profit. Distractor D (60) is the upper break-even point.
A rectangular garden has a fixed perimeter of 100 metres. Its area is given by the function $A(L) = L(50-L)$ , where $L$ is the length of one side in metres. What is the practical domain for the length $L$ in this context?
Answer: $0 < L < 50$ The model is for a physical rectangle. The length of a side, $L$ , must be a positive value, so $L > 0$ . The perimeter is $2L + 2W = 100$ , which simplifies to $L+W=50$ , so the width is $W = 50-L$ . The width must also be a positive value, so $50-L > 0$ , which means $L < 50$ . Combining these two conditions, the practical domain for the length $L$ is $0 < L < 50$ . Distractor A is incomplete as it ignores the constraint on the width. Distractor B is incomplete as it ignores that length must be positive. Distractor D incorrectly restricts the length to be less than or equal to the width (the maximum area occurs at $L=25$ , but lengths up to 50 are physically possible, even if they result in a small area).

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