BC: The least number of cents/km occurs when you get the most km/L of gas. If your tank were near empty, at what speed should you drive to make it to a gas station before you run out? 50 kph I found the solution to this within the previous question: the vertex was equal to 50, 12 therefore there couldn't be anything lower.
FC: Question #6 Fuel Efficiency | By Cy Ryding
1: Q6: The number of cents per kilometer it cost to drive a car depends on how fast you drive it. At low speeds the cost is high because the engine operates inefficiently, while at high speeds the cost is high because the engine must overcome high wind resistance. At moderate speeds cost reaches a minimum. Assume therefore that the number of cents/km varies quadratically with the number of kph. - Suppose that it cost 28, 21, and 16 cents per kilometer to drive at 10, 20, and 30 kph respectively. Write and equation for this function - How much would you spend to drive at 150 kph? - Between which two speeds would must you drive to keep your cost no more than 13 cents/km - Is it possible to spend only 10 cents/km? Justify. -The least number of cents/km occurs when you get the most km/L of gas. If your tank were near empty, at what speed should you drive to make it to a gas station before you run out?
2: Equation for the function: y = .01x^2 - 1x + 37 To find this : 1. I submitted 10, 20, and 30 into the L1 column and 28,21,16 into the L2 column. 2. Next, I wrote QuadReg L1, L2, Y1 and hit enter
3: How much would you spend to drive at 150 kph? $1.12/km I found this by keeping the equation and just searching the table for x = 150. Once i found that all i had to do was look at the corresponding y value.
4: Between which two speeds must you drive to keep under $0.13/km? Between 40 and 60 kph I found this by doing the mirror image of the previous equation. I searched the table for a y value less than/equal to $0.13 and checked the x value, and continued searching until I hit $0.13 again.
5: Is it possible to spend only 10 cents/km? Justify. No, at 50 kph fuel efficiency is at its peak and it is only at $0.12/km I found this by graphing the written equation and hitting trace on the vertex. The vertex was x=50 (50 kph) and y=12 ($0.12)