Universal instruments found that the monthly demand for its new line of Galaxy Home Computers t months after placing the line on the market was given by
D(t) = 2900 − 2300e−0.08t (t > 0)
Graph this function and answer the following questions.
(a) What is the demand after 1 month? After 1 year? After 2 years? after 5 years?
after 1 month computers
after 1 year computers
after 2 years computers
after 5 years computers
(b) At what level is the demand expected to stabilize?
computers
(c) Find the rate of growth of the demand after the tenth month.
computers per month
simply sub in each of the values, I will do the one for 2 years
at 2 years, t = 24
D(24) = 2900- 2300e^(-.08(24))
= 2900-2300e^-1.92
= appr 2563
b) notice that as t gets larger, e^-.08t gets smaller.
So eventually D(t) becomes 2900-0 or 2900
D'(t) = .08(2300)e^-.08t
when t = 10
D'(10) = .08(2300)e^(.08(10)) = 83computers/month
(a) Let's evaluate the demand function D(t) for different values of t.
After 1 month:
D(1) = 2900 - 2300e^(-0.08*1)
Plug in the values and simplify:
D(1) ≈ 2900 - 2300e^(-0.08)
D(1) ≈ 923.07 computers
After 1 year:
D(12) = 2900 - 2300e^(-0.08*12)
Plug in the values and simplify:
D(12) ≈ 2900 - 2300e^(-0.96)
D(12) ≈ 2695.95 computers
After 2 years:
D(24) = 2900 - 2300e^(-0.08*24)
Plug in the values and simplify:
D(24) ≈ 2900 - 2300e^(-1.92)
D(24) ≈ 2834.94 computers
After 5 years:
D(60) = 2900 - 2300e^(-0.08*60)
Plug in the values and simplify:
D(60) ≈ 2900 - 2300e^(-4.8)
D(60) ≈ 2899.99 computers
So, after 1 month, the demand is approximately 923.07 computers.
After 1 year, the demand is approximately 2695.95 computers.
After 2 years, the demand is approximately 2834.94 computers.
After 5 years, the demand is approximately 2899.99 computers.
(b) To find the level at which the demand is expected to stabilize, we can take the limit of the demand function as t approaches infinity.
lim(t→∞) D(t) = lim(t→∞) (2900 - 2300e^(-0.08t))
Since e^(-0.08t) approaches 0 as t approaches infinity, the limit simplifies to:
lim(t→∞) D(t) = 2900 - 2300 * 0
lim(t→∞) D(t) = 2900 computers
Therefore, the demand is expected to stabilize at 2900 computers.
(c) The rate of growth of the demand after the tenth month can be found by taking the derivative of the demand function with respect to time.
D'(t) = 2300 * 0.08e^(-0.08t)
Plug in t = 10 and solve for D'(10):
D'(10) = 2300 * 0.08e^(-0.08*10)
D'(10) ≈ 123.27 computers per month
Therefore, the rate of growth of the demand after the tenth month is approximately 123.27 computers per month.
To answer these questions, we will use the given demand function D(t) = 2900 - 2300e^(-0.08t), where t represents the number of months.
(a) To find the demand after a specific time period, we substitute the value of t into the demand function.
After 1 month:
D(1) = 2900 - 2300e^(-0.08*1)
D(1) = 2900 - 2300e^(-0.08)
D(1) ≈ 2900 - 2300(0.923)
D(1) ≈ 2900 - 2120.9
D(1) ≈ 779.1 computers
After 1 year:
D(12) = 2900 - 2300e^(-0.08*12)
D(12) ≈ 2900 - 2300e^(-0.96)
D(12) ≈ 2900 - 2300(0.383)
D(12) ≈ 2900 - 880.9
D(12) ≈ 2019.1 computers
After 2 years:
D(24) = 2900 - 2300e^(-0.08*24)
D(24) ≈ 2900 - 2300e^(-1.92)
D(24) ≈ 2900 - 2300(0.147)
D(24) ≈ 2900 - 338.1
D(24) ≈ 2561.9 computers
After 5 years:
D(60) = 2900 - 2300e^(-0.08*60)
D(60) ≈ 2900 - 2300e^(-4.8)
D(60) ≈ 2900 - 2300(0.001)
D(60) ≈ 2900 - 2.3
D(60) ≈ 2897.7 computers
(b) To determine the stabilization level of demand, we need to find the limit of the demand function as t approaches infinity.
Taking the limit of D(t) as t approaches infinity:
lim(t→∞) D(t) = lim(t→∞) (2900 - 2300e^(-0.08t))
lim(t→∞) D(t) = 2900 - lim(t→∞) (2300e^(-0.08t))
lim(t→∞) D(t) = 2900 - 2300(0) [since e^(-0.08t) approaches 0 as t approaches infinity]
lim(t→∞) D(t) = 2900
The demand is expected to stabilize at 2900 computers.
(c) The rate of growth of demand after the tenth month can be determined by finding the derivative of the demand function with respect to t and evaluating it at t = 10.
D'(t) = 2300 * 0.08e^(-0.08t)
To find the rate of growth after the tenth month:
D'(10) = 2300 * 0.08e^(-0.08*10)
D'(10) ≈ 2300 * 0.08e^(-0.8)
D'(10) ≈ 2300 * 0.08 * (0.4489)
D'(10) ≈ 82.656 computers per month
Therefore, the rate of growth of the demand after the tenth month is approximately 82.656 computers per month.
To graph the function D(t) = 2900 - 2300e^(-0.08t), you can follow these steps:
1. Determine the range of t values you want to graph. Since t > 0, you may start at t = 1 and go up to a desired time period.
2. Choose a scale for the t-axis and the D-axis. For example, you can use increments of 1 for t and increments of 500 for D.
3. Substitute various values of t into the equation to find the corresponding values of D. Plug in t = 1, t = 12 (1 year), t = 24 (2 years), and t = 60 (5 years).
D(1) = 2900 - 2300e^(-0.08 * 1)
D(12) = 2900 - 2300e^(-0.08 * 12)
D(24) = 2900 - 2300e^(-0.08 * 24)
D(60) = 2900 - 2300e^(-0.08 * 60)
4. Use these values to plot points on the graph. The x-coordinate of each point will be the value of t, and the y-coordinate will be the corresponding value of D.
5. Connect the dots to create a smooth curve on the graph.
(a) The demand after different time periods can be found by substituting the given values of t into the equation.
After 1 month:
D(1) = 2900 - 2300e^(-0.08 * 1) = (calculate the value)
After 1 year:
D(12) = 2900 - 2300e^(-0.08 * 12) = (calculate the value)
After 2 years:
D(24) = 2900 - 2300e^(-0.08 * 24) = (calculate the value)
After 5 years:
D(60) = 2900 - 2300e^(-0.08 * 60) = (calculate the value)
(b) The level at which the demand is expected to stabilize is the long-term behavior of the function as t approaches infinity. As t gets larger and larger, the term e^(-0.08t) approaches 0, leaving only the constant term 2900. Therefore, the demand is expected to stabilize at 2900 computers.
(c) To find the rate of growth of the demand after the tenth month, you can find the derivative dD/dt and evaluate it at t = 10.
dD/dt = -2300 * (-0.08) * e^(-0.08t)
evaluate dD/dt at t = 10 to find the rate of growth in computers per month.