This is urgent! the engine torque y (in foot-pounds) of one model of car is given by y=−3.75x^2+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 b. explain what happens to the engine torque as the speed of the engine increases. as the engine speed increases, the torque until the engine speed reaches thousand revolutions per minute and then the torque

Answer-

a. The engine speed that maximizes torque is 3.1 thousand revolutions per minute  and the maximum torque is 74.68 ft-lbs

b. As the engine speed increases, the torque increases until the engine speed reaches  3.1 thousand revolutions per minute and then the torque decreases.

Solution-

The given equation is,

Where x = the speed of the engine (in thousands of revolutions per minute)

           y = the engine torque y (in foot-pounds)

a.

In order to calculate the max torque and when it is max, we have to take the help of maxima and minima concepts from application of derivatives.

Equating the first derivative to 0, in order to find critical points,

As the value of second derivative is negative, hence the value of the function at x=3.1 will be maximum.

The value of max torque is,

b.

Calculating the roots of the function,

As these are the roots or x-intercept, the value of the function at these points is zero.

As shown in the graph, the function increases from x=0 to x=3.1 (as the speed can not be negative, so neglecting negative values) and after x=3.1 it decreases.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.

66.666666667%

step-by-step explanation:

divide 150 by 100

divide 100 by 1.5

there you go!

decreased by 33.333333334%