a) The value of a computer after t years after purchase is v(t) = 1000e^(-0.45t). At what rate is the computer's value falling after 5 years?

b) Assume that the total revenue received from the sale of x items is given by R(x) = 32 ln (5x+3), while the total cost to produce x items is C(x) = x/4. Find the approximate number of items that should be manufactured so that profit, R(x) - C(x), is a maximum.

a) did you differentiate with respect of t, and then sub in t = 5 ?

b) P(x) = R(x) - C(x)
= 32 ln(5x+3) - x/4

P ' (x) = 32(5/(5x+3)) - 1/4
= 0 for max/min situation

160/(5x+3) = 1/4
5x+3 = 640
etc

a) To find the rate at which the computer's value is falling after 5 years, we need to find the derivative of the function v(t) with respect to t and evaluate it at t = 5.

Step 1: Find the derivative of v(t):
v'(t) = d/dt (1000e^(-0.45t))
= -0.45 * 1000e^(-0.45t) (using the chain rule)

Step 2: Evaluate v'(t) at t = 5:
v'(5) = -0.45 * 1000e^(-0.45*5)
≈ -89.240 (using a calculator to evaluate the exponential function)

Therefore, the computer's value is falling at a rate of approximately -89.240 units per year after 5 years.

b) To find the approximate number of items that should be manufactured to maximize profit, we need to find the critical points of the profit function, R(x) - C(x).

Step 1: Find the profit function:
Profit = R(x) - C(x)
= 32 ln (5x+3) - x/4

Step 2: Find the derivative of the profit function with respect to x:
Profit' = (d/dx)(32 ln (5x+3)) - (d/dx)(x/4)
= 32*(1/(5x+3))*(5) - 1/4
= 16/(5x+3) - 1/4

Step 3: Set the derivative equal to zero and solve for x to find the critical points:
16/(5x+3) - 1/4 = 0

Multiply both sides by 4(5x+3) to eliminate fractions:
16(4) - (5x+3) = 0
64 - 5x - 3 = 0
61 - 5x = 0

Rearrange the equation:
5x = 61
x = 61/5

The approximate number of items that should be manufactured to maximize profit is 12.2 (rounded to one decimal place).