A manufacturer produces bolts of fabric with a fixed width. The quantity q of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can write q = f(p).Then the total revenue earned with selling price p is R(p) = pf(p).Given f(20)=10000 and f '(20) = −350,
find R'(20)
To find R'(20), the derivative of the revenue function R(p) with respect to p evaluated at p=20, we need to use the product rule of differentiation.
The revenue function is given by R(p) = pf(p), where f(p) represents the quantity of fabric sold as a function of the selling price p.
To differentiate the product of two functions, we apply the product rule: (uv)' = u'v + uv', where u and v are differentiable functions.
In this case, u = p and v = f(p). Let's find the derivatives of u and v separately.
The derivative of u = p with respect to p is simply 1.
To find the derivative of v = f(p), we're given that f'(20) = -350. This represents the derivative of f(p) with respect to p evaluated at p=20.
Now, let's differentiate R(p) = pf(p) using the product rule.
R'(p) = (p' * f(p)) + (p * f'(p))
Substituting the values we have:
R'(20) = (1 * f(20)) + (20 * f'(20))
We're given that f(20) = 10000 and f'(20) = -350. Substituting these values into the equation:
R'(20) = (1 * 10000) + (20 * -350)
Simplifying this expression:
R'(20) = 10000 - 7000
R'(20) = 3000
Therefore, the rate of change of the total revenue at a selling price of p=20 is R'(20) = 3000 dollars per yard.