A toy manufacturer has found that its cost,revenue and profit function can be expressed as function of production of toys in a week and C&R&p represent cost,revenue and profit respectively C=6000+2x
R=20x-x^2 all over 2000...suppose production is increasing in the rate of 200 toys per week find the rate of increase in (a)cost (b)revenue (c)profit
To find the rate of increase in the cost, revenue, and profit, given the production is increasing at a rate of 200 toys per week, we need to differentiate the cost, revenue, and profit functions with respect to time.
Given:
Cost function, C = 6000 + 2x
Revenue function, R = (20x - x^2) / 2000
(a) Rate of increase in cost:
To find the rate of increase in cost, we need to differentiate the cost function with respect to time (t).
dC/dt = d(6000 + 2x)/dt
Since the production is increasing at a rate of 200 toys per week, we can substitute dx/dt = 200 into the equation.
dC/dt = 2 * (dX/dt)
Rate of increase in cost = 2 * 200
Therefore, the rate of increase in the cost is 400 units per week.
(b) Rate of increase in revenue:
To find the rate of increase in revenue, we need to differentiate the revenue function with respect to time (t).
dR/dt = d[(20x - x^2) / 2000]/dt
Again, substituting dx/dt = 200 into the equation.
dR/dt = (20 - 2x)/ 2000 * (dx/dt)
Rate of increase in revenue = (20 - 2x) * (dx/dt) / 2000
Since we know that dx/dt = 200, we can substitute it into the equation.
Rate of increase in revenue = (20 - 2x) * 200 / 2000
Therefore, the rate of increase in revenue is given by (20 - 2x) units per week.
(c) Rate of increase in profit:
Profit, P = Revenue - Cost
P = R - C
To find the rate of increase in profit, we need to differentiate the profit function with respect to time (t).
dP/dt = d(R - C)/dt
dP/dt = dR/dt - dC/dt
Substituting the values we obtained from the previous calculations:
dP/dt = (20 - 2x) * 200 / 2000 - 2 * 200
Rate of increase in profit = (20 - 2x) * 200 / 2000 - 400
Therefore, the rate of increase in profit is given by (20 - 2x) units per week minus 400.