i. We can use the Revenue formula to determine what the Revenue equation will be.
As we have polynomials for Price (\(2 + 0.10x\)) and Quantity/People (\(240 000 - 10 000x\)), we can expand these using FOIL to get the final equation:
\(\text{Revenue} = (\text{Price})(\text{Quantity})\)
\(\text{Revenue} = (2 + 0.10x)(240 000 - 10 000x)\)
\(\text{Revenue} = (2)(240 000) + (2)(-10 000x) + (0.10x)(240 000) + (0.10x)(-10 000x)\)
\(\text{Revenue} = 480 000 - 20 000x + 24 000x - 1000x²\)
\(\text{Revenue} = -1000x² + 4000x + 480 000\)
ii. Finding the maximum revenue is similar to finding the vertex for a parabola. In order to do this, we can find the respective roots of the original equations:
\(\text{root}_1: 2 + 0.10x\)
\(\text{root}_1: 0.10x = -2\)
\(\text{root}_1: \cfrac{0.10x}{0.10} = \cfrac{-2}{0.10}\)
\(\text{root}_1: x = -20\)
\(\text{root}_2: 240 000 - 10 000x\)
\(\text{root}_2: 10 000x = 240 000\)
\(\text{root}_2: \cfrac{10 000x}{10 000} = \cfrac{240 000}{10 000}\)
\(\text{root}_2: x = 24\)
Now that we have the roots, we can divide the sum in half to determine the x-coordinate of the vertex. This value represents the # of times the fare should be increased to maximize the revenue:
\(x = \cfrac{-20 + 24}{2}\)
\(x = 2\)
Therefore, we can determine that the fare should get increased 2 times to maximize the revenue.
iii. To determine the new fare, we need to plug the x-coordinate into the Price equation:
\(\text{Price} = 2 + 0.10x\)
\(\text{Price} = 2 + (0.10)(2)\)
\(\text{Price} = 2 + 0.20\)
\(\text{Price} = 2.20\)
Therefore, we can determine that the new fare that maximizes the revenue is \($2.20\).
iv. To determine the # of riders for the maximum revenue, we need to plug the x-coordinate into the Quantity equation:
\(\text{Quantity} = 240 000 - 10 000x\)
\(\text{Quantity} = 240 000 - (10 000)(2)\)
\(\text{Quantity} = 240 000 - 20 000\)
\(\text{Quantity} = 218 000\)
Therefore, we can determine that the # of riders needed for maximum revenue is 218 000.
v. We can determine the maximum revenue by plugging the x-coordinate into the Revenue equation:
\(\text{Revenue} = -1000x² + 4000x + 480 000\)
\(\text{Revenue} = (-1000)(2)² + (4000)(2) + 480 000\)
\(\text{Revenue} = (-1000)(4) + 8000 + 480 000\)
\(\text{Revenue} = -4000 + 8000 + 480 000\)
\(\text{Revenue} = $484 000\)
Therefore, we can determine that the maximum revenue is \($484 000\).