The power, in amps, transmitted by a belt drive from a motor is given by the function P(v)= 100v - 3/16 v^3 .
For what value of v is the power at a maximum value?
Thanks in advance!
At the maximum, P'(v) equals to zero.
P'(v) = 100-(3/16)(2v²) = 0.
Solve for v.
To find the value of v for which the power is at a maximum, we need to differentiate the function P(v) with respect to v and set it equal to zero.
Let's find the derivative of P(v) first:
P(v) = 100v - (3/16)v^3
Differentiating both sides of the equation with respect to v:
dP(v)/dv = 100 - (3/16)(3v^2)
Now, set the derivative equal to zero and solve for v:
100 - (3/16)(3v^2) = 0
Simplifying the equation:
100 - (9/16)v^2 = 0
Rearranging the equation:
(9/16)v^2 = 100
v^2 = (16/9) * 100
v^2 = 1600/9
Taking the square root of both sides:
v = ±√(1600/9)
Therefore, the possible values for v are v ≈ ±14.545.
Since the power being transmitted cannot have a negative value, we can consider v ≈ 14.545 as the only meaningful solution.
Thus, the value of v for which the power is at a maximum is approximately 14.545.
To find the value of ''v'' at which the power is at a maximum, we need to find the critical points of the function P(v). The critical points occur where the derivative of the function is equal to zero or undefined.
The first step is to find the derivative of P(v) with respect to ''v''.
P(v) = 100v - (3/16)v^3
To find the derivative, we need to apply the power rule.
dP/dv = d(100v)/dv - d((3/16)v^3)/dv
dP/dv = 100 - (3/16)(3v^2)
Simplifying further,
dP/dv = 100 - (9/16)v^2
Now, we set this derivative equal to zero and solve for ''v'' to find the critical points:
100 - (9/16)v^2 = 0
(9/16)v^2 = 100
v^2 = (100 * 16) / 9
v^2 = 1600 / 9
Taking the square root of both sides,
v = ± sqrt(1600 / 9)
v ≈ ±14.37
Therefore, at v ≈ 14.37 or -14.37, the power is at a maximum.
Note: Since power cannot be negative, the value of 'v' will be approximately 14.37 for the maximum power.