Math 2 Quiz: Modeling With Linear Quadratic Systems
6 questions · exam conditions
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Modeling With Linear Quadratic SystemsQuestion 1 of 6

A company's daily production cost is C(x) = x² + 40x + 500 dollars for x units. Their daily revenue is R(x) = 100x - 2x² dollars. On a day when they want to break even (profit = 0), which equation correctly represents this condition?

3x² - 60x + 500 = 0
-3x² + 60x - 500 = 0
x² + 40x + 500 = 100x - 2x²
-x² - 140x - 500 = 0
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Math 2 Quiz

Math 2 Quiz: Modeling With Linear Quadratic Systems

Practice Modeling With Linear Quadratic Systems in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Modeling With Linear Quadratic Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A company's daily production cost is C(x) = x² + 40x + 500 dollars for x units. Their daily revenue is R(x) = 100x - 2x² dollars. On a day when they want to break even (profit = 0), which equation correctly represents this condition?

  1. 3x² - 60x + 500 = 0 (correct answer)
  2. -3x² + 60x - 500 = 0
  3. x² + 40x + 500 = 100x - 2x²
  4. -x² - 140x - 500 = 0
Explanation: Break-even occurs when Profit = Revenue - Cost = 0, so R(x) = C(x). This gives us 100x - 2x² = x² + 40x + 500. Rearranging: 100x - 2x² - x² - 40x - 500 = 0, which simplifies to -3x² + 60x - 500 = 0. Multiplying by -1 gives 3x² - 60x + 500 = 0. Choice B is the equation before multiplying by -1. Choice C shows the condition but isn't simplified. Choice D has sign errors.

Question 2

A particle moves along a path where its vertical position yy (in meters) is given by y(t)=t2−6t+13y(t) = t^2 - 6t + 13 and its horizontal position follows x(t)=2t+1x(t) = 2t + 1, where tt is time in seconds. At what horizontal position will the particle reach its minimum vertical position?

  1. x=5x = 5 meters
  2. x=6x = 6 meters
  3. x=7x = 7 meters (correct answer)
  4. x=8x = 8 meters
Explanation: To find the minimum vertical position, we need to minimize y(t)=t2−6t+13y(t) = t^2 - 6t + 13. This is a parabola opening upward, so the minimum occurs at the vertex. The vertex of at2+bt+cat^2 + bt + c occurs at t=−b2a=−−62(1)=3t = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3. At t=3t = 3, the horizontal position is x(3)=2(3)+1=7x(3) = 2(3) + 1 = 7 meters. We can verify: y(3)=9−18+13=4y(3) = 9 - 18 + 13 = 4, which is indeed the minimum value since y(t)=(t−3)2+4y(t) = (t-3)^2 + 4. Choice A (5 meters) corresponds to t=2t = 2. Choice B (6 meters) corresponds to t=2.5t = 2.5. Choice D (8 meters) corresponds to t=3.5t = 3.5. All of these times give higher vertical positions than t=3t = 3.

Question 3

A farmer wants to fence a rectangular garden against an existing wall. The wall will serve as one side, so fencing is needed for only three sides. If the farmer has 120 feet of fencing and wants to maximize the area, the optimal dimensions create an area described by A=x(120−2x)A = x(120-2x), where xx is the width perpendicular to the wall. If the farmer decides to reduce the total fencing used to exactly 100 feet while maintaining the same rectangular design, what will be the new maximum area?

  1. 1200 square feet
  2. 1250 square feet (correct answer)
  3. 1300 square feet
  4. 1350 square feet
Explanation: With 100 feet of fencing, the new area function becomes A=x(100−2x)A = x(100-2x), where xx is the width perpendicular to the wall. To maximize, we find the vertex of this parabola. The area function A(x)=100x−2x2A(x) = 100x - 2x^2 is maximized when dAdx=100−4x=0\frac{dA}{dx} = 100 - 4x = 0, giving x=25x = 25. The length parallel to the wall is 100−2(25)=50100 - 2(25) = 50 feet. Therefore, the maximum area is 25×50=125025 \times 50 = 1250 square feet. Choice A (1200) might result from calculation errors. Choice C (1300) could come from using the wrong fencing total. Choice D (1350) might result from incorrectly applying the original 120-foot constraint.

Question 4

A small business manufactures and sells custom phone cases. The daily production cost (in dollars) is given by C(x) = 2x² + 50x + 200, where x is the number of cases produced per day. The daily revenue (in dollars) is R(x) = 120x - 0.5x².

The business wants to determine when their daily profit equals $800. Which system of equations correctly models this situation?

  1. P(x) = -2.5x² + 70x - 200 and P(x) = 800 (correct answer)
  2. P(x) = -2.5x² + 70x - 200 and P(x) = -800
  3. P(x) = 1.5x² - 170x + 200 and P(x) = 800
  4. P(x) = 2.5x² - 70x + 200 and P(x) = 800
Explanation: Profit = Revenue - Cost, so P(x) = R(x) - C(x) = (120x - 0.5x²) - (2x² + 50x + 200) = -2.5x² + 70x - 200. To find when profit equals $800, we set P(x) = 800. Choice B incorrectly uses -800. Choice C has wrong signs in the quadratic coefficient. Choice D incorrectly adds costs instead of subtracting.

Question 5

A company's monthly profit P(x) (in thousands) depends on advertising spend x (in thousands) according to P(x) = -x² + 12x - 20. They want to find when their profit equals their advertising budget. Which equation represents this condition?

  1. -x² + 12x - 20 = x (correct answer)
  2. -x² + 12x - 20 = -x
  3. -x² + 11x - 20 = 0
  4. -x² + 13x - 20 = 0
Explanation: When profit equals advertising budget, we have P(x) = x, which gives us -x² + 12x - 20 = x. Choice B incorrectly uses -x for the budget. Choices C and D represent the simplified forms but with errors: C would come from incorrectly moving x to the left side as +x instead of -x, while D would come from incorrectly moving x as -x instead of +x.

Question 6

A stone is dropped from a bridge 144 feet above a river. Its height is h(t) = -16t² + 144 feet after t seconds. At the same time, a fish jumps from the water following the path h(t) = -16t² + 32t.

How many seconds after the stone is dropped will it be at the same height as the jumping fish?

  1. 1.5 seconds and 3 seconds
  2. 2.25 seconds only
  3. 4.5 seconds only (correct answer)
  4. No solution exists
Explanation: Set the heights equal: -16t² + 144 = -16t² + 32t. Simplifying: 144 = 32t, so t = 144/32 = 4.5 seconds. The -16t² terms cancel out, leaving a linear equation with one solution. Choice A suggests two solutions, which would occur if we had a quadratic equation. Choice B gives an incorrect calculation. Choice D incorrectly suggests no solution exists.