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
To find the rate of change of the sales volume when t = 1, we need to calculate the derivative of the sales function, S(t).
The given sales function is S(t) = 20,000(1 + e^(-0.6t)), where t represents the number of days after the end of the sales promotion.
To find the derivative, we can use the chain rule, which states that if we have a function of the form f(g(t)), the derivative of f(g(t)) with respect to t is f'(g(t)) * g'(t).
In this case, let's consider f(u) = 20,000(1 + e^u) and g(t) = -0.6t.
First, let's take the derivative of f(u) with respect to u. The derivative of 20,000(1 + e^u) is 20,000 * e^u.
Next, let's find the derivative of g(t) with respect to t. The derivative of -0.6t is -0.6.
Now, let's apply the chain rule by multiplying the derivatives f'(g(t)) * g'(t).
f'(g(t)) = 20,000 * e^(g(t)) = 20,000 * e^(-0.6t).
g'(t) = -0.6.
Therefore, the derivative of S(t) is given by S'(t) = f'(g(t)) * g'(t) = 20,000 * e^(-0.6t) * (-0.6).
To find the rate of change of sales volume at t = 1, plug in t = 1 into the derivative function S'(t):
S'(1) = 20,000 * e^(-0.6 * 1) * (-0.6) = 20,000 * e^(-0.6) * (-0.6).
To get the answer to the nearest dollar, we need to evaluate this expression. Using a calculator or a math software:
S'(1) ≈ -8910.
Therefore, the rate of change of sales volume when t = 1 is approximately -8910 sales per day (rounded to the nearest dollar).