Model Airplane Radio
Calculus Problem for the few?
the average worker at wakefield avionics can assemble N(t) = -2t^3 + 12t^2 + 2t
(0 <= t <= 4)
ready to fly radio controlled model airplanes t hr into the 8 am to 12 noon morning shift. at what time during this shift is the average woker performing at peak efficiency? Thanks for your time
peak efficiency is actually at 10 am, at least that is what the books answer is
N(t) = -2t^3 + 12t^2 + 2t
N’(t) = -6t^2+24t+2 =0
-6t^2+24t+2=0
This equation is of form ax^2+bx+c
a = -6 b = 24 c = 2
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-24 +/-sqrt(24^2-4(-6)(2)]/(2)(-6)
discriminant is b^2-4ac =624
x=[-24 +√(624)] / (2)(-6)
x=[-24 -√(624)] / (2)(-6)
x=[-24+24.979991993593593] / -12
x=[-24-24.979991993593593] / -12
The roots are -0.0817 and 4.0817
The maximum occurs when t=-0.08 ot t=4.08 which both do not fall in the interval (0,4)
If you graph the function, the function is maximum at t=4.08
So, assume 4.08=4
At 12:00 p.m., the average woker performing at peak efficiency.
d^2N/dt^2 = -12t+24 <0, when t=4
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