a woman's clothing chain store, found that t days after the end of a sales promotion the volume of sales was given by S(t)= 20,000 (1+e^-.6t), 0 less than or equal to t less than or equal to 5 dollars. find to the nearest dollars the rate of change of sales volume when t=1

The rate of change of sales volume is the derivative of S(t)

S'(t) = dS/dt = 20,000 * -0.6 * e^(-0.6*t)

Let t = 1, and find the value of S'(1)

To find the rate of change of sales volume when t=1, we need to find the derivative of the sales function S(t) with respect to t and evaluate it at t=1.

First, let's find the derivative of S(t). The given sales function is S(t) = 20,000(1+e^(-0.6t)). To find its derivative, we can apply the chain rule.

1. Start by differentiating the outer function, which is constant 20,000, with respect to t. The derivative of a constant is zero, so we can ignore it for now.

2. Now differentiate the inner function, which is (1+e^(-0.6t)). The derivative of 1 with respect to t is zero, so we can ignore it as well.

3. To differentiate e^(-0.6t), we can use the chain rule. The derivative of e^u with respect to u is e^u multiplied by the derivative of u with respect to t.

In this case, u is -0.6t, so the derivative of e^(-0.6t) with respect to t is e^(-0.6t) multiplied by the derivative of -0.6t with respect to t.

Since the derivative of -0.6t with respect to t is -0.6, we can simplify the expression to -0.6e^(-0.6t).

Now, let's put it all together:

S'(t) = 20,000(0 + -0.6e^(-0.6t))
= -12,000e^(-0.6t)

To find the rate of change of sales volume when t=1, we substitute t=1 into the derivative S'(t):

S'(1) = -12,000e^(-0.6(1))
= -12,000e^(-0.6)

Now we can calculate the value to the nearest dollar:

S'(1) ≈ -12,000e^(-0.6) ≈ -12,000(0.548812) ≈ -6,585

Therefore, the rate of change of sales volume when t=1 is approximately -6,585 dollars per day.