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,
[tex]y=-3.75x^2+23.2x+38.8[/tex]
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.
[tex]\Rightarrow y=-3.75x^2+23.2x+38.8[/tex]
[tex]\Rightarrow {y}'=-7.5x+23.2[/tex]
[tex]\Rightarrow {y}''=-7.5[/tex]
Equating the first derivative to 0, in order to find critical points,
[tex]\Rightarrow {y}'=0[/tex]
[tex]\Rightarrow -7.5x+23.2=0[/tex]
[tex]\Rightarrow x=\frac{-23.2}{-7.5} = 3.093 \approx 3.1[/tex]
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,
[tex]y(3.1)=-3.75(3.1)^2+23.2(3.1)+38.8=74.682 \approx 74.68[/tex]
b.
Calculating the roots of the function,
[tex]\Rightarrow y=0[/tex]
[tex]\Rightarrow -3.75x^2+23.2x+38.8=0[/tex]
[tex]x=-1.369,7.556[/tex]
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.
