A car travels at a constant speed and burns g(x) litres of gas per kilometer, where x is the speed of the car in kilometers per hour and g(x) = 1280 + x^2/320x
a) If gas costs $1.29 per litre, find the cost function C(x) that expresses the cost of the fuel for a 200 km trip as a function of the speed.
b) What driving speed will make the cost of fuel equal to $300?
c) Determine the driving speed that will minimize the total fuel cost for the trip.
a) To find the cost function, C(x), we need to multiply the gas consumption, g(x), by the cost per litre, which is $1.29. So, the cost function C(x) can be expressed as:
C(x) = g(x) * 1.29
Since g(x) = 1280 + (x^2/320x), we substitute it into the cost function:
C(x) = (1280 + (x^2/320x)) * 1.29
Simplifying further, we have:
C(x) = 1651.2 + (x/247)
Therefore, the cost function C(x) that expresses the cost of the fuel for a 200 km trip as a function of the speed is C(x) = 1651.2 + (x/247).
b) To find the driving speed that will make the cost of fuel equal to $300, we set C(x) equal to 300 and solve for x:
1651.2 + (x/247) = 300
Subtracting 1651.2 from both sides:
x/247 = 300 - 1651.2
x/247 = -1351.2
Multiplying both sides by 247:
x = -1351.2 * 247
Well, that's a negative speed! I guess you can consider yourself a superhero, going back in time while driving.
c) To determine the driving speed that will minimize the total fuel cost for the trip, we need to find the minimum point on the cost function, C(x). Since C(x) = 1651.2 + (x/247), we can take the derivative of C(x) with respect to x and set it equal to zero to find the critical point:
C'(x) = 1/247 = 0
But wait, that derivative is a constant! This means that the cost function is constant, and the cost will be the same regardless of the driving speed. So the driving speed that will minimize the total fuel cost for the trip is any speed you like! Go as fast or as slow as you want, and the cost will remain the same. Just try not to get a speeding ticket while enjoying the cost-saving benefits!
a) To find the cost function C(x), we need to calculate the total amount of gas burned during the 200 km trip and multiply it by the cost per litre.
The amount of gas burned per kilometer is given by g(x) = 1280 + x^2/(320x).
To find the total amount of gas burned during the 200 km trip, we integrate g(x) over the interval [0, 200]:
∫(1280 + x^2/(320x)) dx.
Simplifying the integral:
∫(1280 + x/(320)) dx = 1280x + (1/320) ln|x| + C.
Since we are considering a specific trip from 0 to 200 km, we can use the definite integral:
∫[0, 200] (1280 + x/(320)) dx = [1280x + (1/320) ln|x|] evaluated from 0 to 200.
Substituting the limits of integration:
[1280(200) + (1/320) ln|200|] - [1280(0) + (1/320) ln|0|].
As ln(0) is undefined, the second term can be disregarded, and we have:
C(x) = 1280(200) + (1/320) ln|200|.
Therefore, the cost function C(x) that expresses the cost of fuel for a 200 km trip as a function of the speed is:
C(x) = 256,000 + (1/320) ln|200|.
b) To find the driving speed that will make the cost of fuel equal to $300, we need to solve the equation C(x) = 300.
Setting C(x) = 300:
256,000 + (1/320) ln|200| = 300.
Simplifying the equation:
(1/320) ln|200| = 300 - 256,000.
Divide both sides by (1/320):
ln|200| = 320(300 - 256,000).
Taking the exponential of both sides to eliminate the natural logarithm:
|200| = e^(320(300 - 256,000)).
Since the exponential of a positive number is always positive, we can drop the absolute value signs:
200 = e^(320(300 - 256,000)).
To solve for x, we need to take the natural logarithm of both sides:
ln(200) = 320(300 - 256,000).
Using a calculator to evaluate ln(200), we find:
5.2983 = 320(300 - 256,000).
Dividing both sides by 320:
5.2983/320 = 300 - 256,000.
Simplifying:
0.01655 = 300 - 256,000.
Rearranging the equation to solve for x:
x = (300 - 256,000 + 0.01655) / 0.01655.
Evaluating this expression, we find:
x ≈ 12,353.33 km/h.
Therefore, the driving speed that will make the cost of fuel equal to $300 is approximately 12,353.33 km/h.
c) To determine the driving speed that will minimize the total fuel cost for the trip, we need to minimize the cost function C(x).
Differentiating C(x) with respect to x:
C'(x) = 0 + (1/320) (1/x) = (1/320x).
Setting C'(x) equal to zero:
(1/320x) = 0.
Since the derivative is never zero, we cannot find a minimum by setting the derivative equal to zero. However, we can determine the limits of C(x) as x approaches infinity and x approaches zero to guide our analysis.
As x approaches infinity, C(x) approaches infinity.
As x approaches zero, C(x) approaches infinity.
Therefore, the total fuel cost increases as the driving speed approaches both zero and infinity, but there is no minimum fuel cost.
Hence, there is no driving speed that will minimize the total fuel cost for the trip.
To answer these 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 x.
a) To find the cost function, we can multiply the amount of gas burned per kilometer (g(x)) by the distance traveled (200 km).
So, the cost function C(x) = g(x) * 200
Now, let's substitute g(x) into the cost function:
C(x) = (1280 + (x^2/320)) * 200
Simplifying this equation will give us the cost of the fuel for a 200 km trip as a function of the speed 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.
So, we have the equation 300 = (1280 + (x^2/320)) * 200.
Solving this equation will give us the driving speed that corresponds to a fuel cost of $300.
c) To determine the driving speed that will minimize the total fuel cost for the trip, we need to find the minimum point of the cost function C(x).
To find the minimum point, we can take the derivative of the cost function C(x) with respect to x, set it equal to zero, and solve for x.
First, let's find the derivative of C(x):
C'(x) = (1280 + (x^2/320))' * 200
= (x^2/320)' * 200
= (2x/320) * 200
= x/8
Setting the derivative equal to zero:
x/8 = 0
Since x/8 is only equal to zero when x = 0, we have a minimum point at x = 0.
Therefore, the driving speed that will minimize the total fuel cost for the trip is 0 km/h.