Ralph is taking his final exam. After t minutes into the exam Ralph has correctly solved P(t)=-.001t^3+.078t^2+.16t problems.
a) at what time was Ralph solving problems most efficiently?
b) at what time did Ralph's efficiency decline?
dP/dt is how many problems he solves per minute
the maximum of dP/dt is where he is most efficient
I assume that after he is most efficient, he gets less efficient.
now
speed = v = dP/dt = -.003t^2+ .156 t +.16
dv/dt = 0 for max or min
0 = -.006t + .156
t = 26 for max or min efficiency
Now is it max or min or neurtral?
If the next derivative is negative, we were at a maximum
the next derivative is -.006
so indeed we have a maximum at t = 26, and it is all downhill from there.
To find at what time Ralph was solving problems most efficiently, we need to find the highest point on the graph of P(t). This can be done by finding the maximum value of P(t).
a) To find the time at which Ralph was solving problems most efficiently:
1. Differentiate P(t) with respect to t to find its derivative: P'(t).
P'(t) = dP(t)/dt = -0.003t^2 + 0.156t + 0.16.
2. Set P'(t) equal to zero and solve for t to find the critical points:
-0.003t^2 + 0.156t + 0.16 = 0
3. Solve the quadratic equation for t by factoring or using the quadratic formula.
4. The values of t obtained from the solutions to the quadratic equation represent the time at which Ralph was solving problems most efficiently.
b) To find the time at which Ralph's efficiency declined:
1. The efficiency is declining when the slope of the graph is negative or decreasing.
2. Analyze the concavity of the graph by finding the second derivative of P(t): P''(t).
P''(t) = d^2P(t)/dt^2 = -0.006t + 0.156.
3. Set P''(t) equal to zero and solve for t to find any inflection points.
4. Solve the equation -0.006t + 0.156 = 0 to find the time at which Ralph's efficiency declined.
The critical points obtained from solving these equations will give us the answers to both parts of the question.