The engine torque y (in foot-pounds) of one model of car is given by y=−3.75x2+23.2x+38.8, where x is the speed (in thousands of revolutions per minute) of the engine. a. To the nearest ten, what is the engine speed that maximizes torque? revolutions per minute To the nearest hundredth, what is the maximum torque? ft-lbs Question 2

Step-by-step explanation:

If the engine torque y (in foot-pounds) of one model of car is given by y=−3.75x^2+23.2x+38.8

The engine speed is at maximum if dy/dx = 0

dy/dx = -2(3.75)x+23.2

dy/dx = -7.5x + 23.2

since dy/dx = 0

0 =  -7.5x + 23.2

7.5x = 23.2

x = 23.2/7.5

x = 3.093

Hence the maximum torque is 3.09 rev/min

4.6875 pounds

step-by-step explanation:

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