The number of points that john has earned in a game varies every second according to the formula p=t^4-4t^3-18t^2+4t+18 where p is the number of points and t is the time that has elapsed in seconds. When is john earning points at the slowest rate possible.
How would i go about finding the minimum in this question
working and solutions please
Anonymous
So the question is asking about when john is earning points at the slowest rate possible. If P(t) represents the number of points he gets with regards to time, P'(t)= the rate that john is achieving points.
P'(t)= 4t^3-12t^2+32t+4
P'(t)=4(t^3-3t^2+8t+1)
Now you need to find when P'(t) is at it's lowest value. We see P(t) is concave up and a parabola so we know that any critical points is going to be a minimum.
Set P'(t)=0 and solve
david
P = t^4-4t^3-18t^2+4t+18
P ' = 4t^3 - 12t^2 - 36t + 4 << derivative .. set = 0
4t^3 - 12t^2 - 36t + 4 = 0 << divide by 4
t^3 - 3t^2 - 9t + 1 = 0 << factor
1st attempt: (t^3 - 1) - 3t(t + 9 ) << not working
2nd attempt: (t^3 - 9t) - (3t^2 -1) <<< not working
3rd : (t^3 - 3t^2) - (9t - 1) <<< no ... I do not think it factors under normal methods using integers ...
you should graph the derivative to find its zeros ... where it crosses the x axis .. then use those to find the min. value.
cidyah
p=t^4-4t^3-18t^2+4t+18
dp/dt = 4t^3-12t^2-36t+4
we want the t when dp/dt is minimum
d^2p/dt^2 = 12t^2-24t-36 = 0
12(t^2-2t-3) = 0
12(t-3)(t+1) = 0
t = 3
d^3p/dt^3 = 24t -24
when t=3 , d^3p/dt^3 > 0, so dp/dt is minimum when t=3 sec
when t=3, dp/dt = -104
Anonymous
yes