a car travels at a constants speed and uses G(x) litres of gas per kilomotre, where x is the speed of the car in kilomotres oer hour and G(x)=(1280+x^2)/(320x)
a>if fuel costs 1.29 per litre, determine the cost function C(x)that expresses the cost of the fuel for a 200KM trip as a function of the speed.
b>what driving speed will make the cost of fuel equal to 300
c>what driving speed will minimize the cost of fuel for the trip?
a> Well, to find the cost function C(x) that expresses the cost of fuel for a 200KM trip as a function of the speed, we just need to multiply the amount of fuel used (G(x)) by the cost per liter (1.29) and by the distance traveled (200KM). So, C(x) = G(x) * 1.29 * 200.
b> Now, to find the driving speed that will make the cost of fuel equal to 300, we just need to solve the equation C(x) = 300. That means G(x) * 1.29 * 200 = 300. Sounds like a job for some algebraic fun!
c> Lastly, to find the driving speed that will minimize the cost of fuel for the trip, we need to find the minimum point of the cost function C(x). We can do this by taking the derivative of C(x) and finding where it equals zero. And remember, even in the world of math, minimizing cost is always a good thing. It's like getting a discount on your fizzy clown shoes!
a> To find the cost function C(x) that expresses the cost of fuel for a 200KM trip as a function of the speed, we need to multiply the amount of fuel consumed (in liters) by the cost per liter.
The amount of fuel consumed for a 200KM journey at a speed of x kilometers per hour is given by G(x) * 200. Therefore, the cost function C(x) can be expressed as:
C(x) = G(x) * 200 * 1.29
Now let's substitute the expression for G(x):
C(x) = (1280 + x^2) / (320x) * 200 * 1.29
Simplifying, we get:
C(x) = (2.58 * (1280 + x^2)) / x
Therefore, the cost function C(x) that expresses the cost of fuel for a 200KM trip as a function of the speed is (2.58 * (1280 + x^2)) / x.
b> To find the driving speed that will make the cost of fuel equal to 300, we can set the cost function C(x) equal to 300 and solve for x.
C(x) = 300
(2.58 * (1280 + x^2)) / x = 300
Multiplying both sides by x:
2.58 * (1280 + x^2) = 300 * x
Expanding the expression:
2.58 * 1280 + 2.58 * x^2 = 300 * x
Rearranging the equation:
2.58 * x^2 - 300 * x + 2.58 * 1280 = 0
Now we can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2.58, b = -300, and c = 2.58 * 1280.
Solving this equation will give us the driving speed that will make the cost of fuel equal to 300.
c> To find the driving speed that will minimize the cost of fuel for the trip, we need to find the critical points of the cost function C(x), where the derivative of C(x) is equal to zero.
To do this, we will take the derivative of the cost function C(x) with respect to x and set it equal to zero.
C'(x) = 0
We will differentiate the cost function C(x) by applying the quotient rule:
C'(x) = [(2.58 * x * (1280 + x^2)) - ((2.58 * (1280 + x^2)) * 1)] / x^2
Simplifying and setting the derivative equal to zero:
[(2.58 * x * (1280 + x^2)) - (2.58 * (1280 + x^2))] / x^2 = 0
(2.58 * x * (1280 + x^2) - 2.58 * (1280 + x^2)) / x^2 = 0
Multiplying through by x^2:
2.58 * x * (1280 + x^2) - 2.58 * (1280 + x^2) = 0
Expanding and rearranging the equation:
2.58 * x^3 + 2.58 * 1280 * x - 2.58 * 1280 = 0
Now we can solve this cubic equation for x using numerical methods or approximation techniques to find the driving speed that will minimize the cost of fuel for the trip.
To answer the questions, we need to find the cost function C(x) that expresses the cost of fuel for a 200 km trip as a function of the speed, determine the driving speed that will make the cost of fuel equal to 300, and find the driving speed that will minimize the cost of fuel for the trip.
a) To find the cost function C(x), we need to calculate the total fuel used for a 200 km trip and then multiply it by the cost of fuel per liter.
First, let's find the fuel used for a 200 km trip. We know that the fuel usage is given by G(x) liters per kilometer, so the total fuel used for 200 km is:
Fuel used = G(x) * Distance = G(x) * 200 km
Next, we calculate the cost of the fuel using the fuel used and the cost per liter:
Cost of fuel = Fuel used * Cost per liter = G(x) * 200 km * 1.29
Therefore, the cost function C(x) is:
C(x) = 1.29 * (G(x) * 200 km)
To simplify further, substitute the given formula for G(x) into the cost function:
C(x) = 1.29 * (G(x) * 200 km)
= 1.29 * ((1280 + x^2) / (320x)) * 200 km
Simplifying further:
C(x) = 2.0325 * ((1280 + x^2) / x) km
So, the cost function C(x) is 2.0325 * ((1280 + x^2) / x) km.
b) To find the driving speed that will make the cost of fuel equal to 300, we set the cost function C(x) equal to 300 and solve for x:
2.0325 * ((1280 + x^2) / x) = 300
Simplifying the equation and solving for x may involve rearranging the equation and applying algebraic operations. Please note that the equation may be non-linear, so it might require the use of a numerical method like a graphing calculator or an online equation solver to find the precise value of x.
c) To find the driving speed that will minimize the cost of fuel for the trip, we need to find the value of x that minimizes the cost function C(x). We can do this by finding the derivative of C(x) with respect to x, setting it equal to zero, and solving for x.
Differentiating C(x) with respect to x:
C'(x) = 2.0325 * (1280(x) - (1280 + x^2)) / x^2
Setting C'(x) equal to zero and solving for x may involve algebraic manipulation and simplification. Once again, keep in mind that the solution might require the use of a numerical method.
G(x) = 4/x + x/320
(a) C(x) = cost for x km trip
$ = liters * $/liter
liters = L/km * km
C(x) = G(x) * 200 * 1.29
= 258*G(x)
= 258(4/x + x/320) = 1032/x + 258/320 * x
(b) 300 = 1032/x + 258/320 x
96000x = 330240 + 258x^2
x = 3.4 or 368.6 ???!!!??
(c) C' = -1032/x^2 + 0.80625
C' = 0 at x = 35.8 km/h
If my math is right, that's some weird numbers...