- A particle moves along a path where its vertical position y (in meters) is given by y(t)=t2−6t+13 and its horizontal position follows x(t)=2t+1, where t is time in seconds. At what horizontal position will the particle reach its minimum vertical position?
Answer: x=7 meters
To find the minimum vertical position, we need to minimize y(t)=t2−6t+13. This is a parabola opening upward, so the minimum occurs at the vertex. The vertex of at2+bt+c occurs at t=−2ab=−2(1)−6=3. At t=3, the horizontal position is x(3)=2(3)+1=7 meters. We can verify: y(3)=9−18+13=4, which is indeed the minimum value since y(t)=(t−3)2+4. Choice A (5 meters) corresponds to t=2. Choice B (6 meters) corresponds to t=2.5. Choice D (8 meters) corresponds to t=3.5. All of these times give higher vertical positions than t=3. - The business wants to determine when their daily profit equals $800. Which system of equations correctly models this situation?
Answer: P(x) = -2.5x² + 70x - 200 and P(x) = 800
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.
- 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), where x 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?
Answer: 1250 square feet
With 100 feet of fencing, the new area function becomes A=x(100−2x), where x is the width perpendicular to the wall. To maximize, we find the vertex of this parabola. The area function A(x)=100x−2x2 is maximized when dxdA=100−4x=0, giving x=25. The length parallel to the wall is 100−2(25)=50 feet. Therefore, the maximum area is 25×50=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. - How many seconds after the stone is dropped will it be at the same height as the jumping fish?
Answer: 4.5 seconds only
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.
- 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?
Answer: 3x² - 60x + 500 = 0
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.
- 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?
Answer: -x² + 12x - 20 = x
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.
