the function f(x)= -o.005t^2+0.8t+12 models the relationship between a certain vehicle’s speed and fuel economy, where is the fuel economy in kilometres per litre and x is the speed of the vehicle in kilometres per hour. Determine the rate of change in fuel economy for 10 km/h intervals in speed, and use your results to determine the speed that gives the best fuel economy.
ANSWER : 18km/h
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To determine the rate of change in fuel economy for 10 km/h intervals in speed, we need to find the derivative of the given function with respect to x (speed). The derivative represents the rate of change of the function.
Let's find the derivative of the function f(x) = -0.005t^2 + 0.8t + 12.
The derivative of -0.005t^2 is -0.01t, using the power rule.
The derivative of 0.8t is 0.8, using the constant rule.
The derivative of 12 is 0, since it is a constant.
So, the derivative of f(x) is f'(x) = -0.01t + 0.8.
Now, we can use this derivative to calculate the rate of change in fuel economy for 10 km/h intervals in speed. We will evaluate the derivative at various speeds:
For a speed of 10 km/h:
f'(10) = -0.01(10) + 0.8 = 0.8 - 0.1 = 0.7 km/L per 10 km/h.
For a speed of 20 km/h:
f'(20) = -0.01(20) + 0.8 = 0.8 - 0.2 = 0.6 km/L per 10 km/h.
For a speed of 30 km/h:
f'(30) = -0.01(30) + 0.8 = 0.8 - 0.3 = 0.5 km/L per 10 km/h.
We can continue this process for other 10 km/h intervals. The rate of change in fuel economy is decreasing as the speed increases.
To determine the speed that gives the best fuel economy, we need to find where the rate of change in fuel economy is equal to zero or closest to zero. In this case, we are looking for a minimum.
We have the equation f'(x) = -0.01t + 0.8, and we want to find where f'(x) = 0.
-0.01t + 0.8 = 0
-0.01t = -0.8
t = -0.8 / -0.01
t = 80 km/h.
Therefore, the speed that gives the best fuel economy is 80 km/h. However, it is worth noting that this is a quadratic function, and the given interval might not include the full range of the function. So, it's essential to consider the specific range provided in the problem.