Question

Assume that the profit generated by a product is given by P(x) = 3 x, where x is the number of units sold. If the profit keeps changing at a rate of $1000 per month, then how fast are the sales changing when the number of units sold is 600? (Round your answer to the nearest dollar per month.)

Answer 1

The sales are changing at a rate of approximately $333 per month when 600 units are sold, based on the given profit function P(x) = 3x and a constant profit change of $1000 per month.

Step-by-step explanation:

Rate of Sales Change from Profit Function

The profit function for a product is given as P(x) = 3x, where x is the number of units sold, and P(x) represents the profit. If the profit changes at a constant rate of $1000 per month, we are interested in determining how fast the sales are changing when 600 units are sold. The rate of sales change can be represented by the derivative of the profit function with respect to time (t), which gives us the rate at which units are sold per month.

We know that: dP/dt = 3 * dx/dt
Since the profit is changing at $1000/month, we have: 1000 = 3 * dx/dt
Solving for dx/dt, we find that dx/dt = 1000 / 3 = $333.33/month (approximately).

Therefore, when the number of units sold is 600, the sales are changing at an approximate rate of $333 per month.

Answer 2

D(x) / dt ≈ - 10⁻⁷

Step-by-step explanation: Incomplete Question from google I found that the expression P(x) = 3 x is incorrect. The expression should be 3/√x

We assume :

P(x) = 3 / √x (1)

We know that DP(x)/dt = 1000 $/month

and

Differentiating on both sides of the equation (1)

DP(x)/dt = 3 * -1/2 * D(x)/dt /x√x ⇒ DP(x)/dt =[ -3/2x√x ] D(x)/dt

To evaluate how fast are the sales changing when the number f units (x) is 600

DP(x)/dt =[ -3/2x√x ] D(x)/dt

1000 = [ -3/2*600*√600 ] D(x) /dt

D(x) / dt = - 1000/1.02*10⁻⁴

D(x) / dt ≈ - 10⁻⁷