Question

A certain company determines that the rate of change of monthly profit p, as a function of monthly advertising expenditure x, is proportional to the difference between a maximum amount of $11,000 and p. furthermore, if no money is spent on advertising, the profit is $1,000; if $100 is spent on advertising, the profit is $6,000.

write and solve a suitable initial-value problem to determine the profit if $200 were spent on advertising.

Answer 1

To solve the initial-value problem and determine the profit if $200 were spent on advertising, we can set up a differential equation using the given information. By finding the constant of proportionality and solving the initial-value problem, we can find the value of the monthly profit. The detailed solution provides step-by-step explanations.

Step-by-step explanation:

To solve this problem, we need to set up an initial-value problem using the given information. Let's denote the monthly profit as p and the monthly advertising expenditure as x.

According to the problem, the rate of change of monthly profit p, as a function of monthly advertising expenditure x, is proportional to the difference between a maximum amount of $11,000 and p. This can be expressed as:

dp/dx = k(11000 - p)

where k is the constant of proportionality.

We also know that when no money is spent on advertising (x = 0), the profit is $1,000 (p = 1000). This gives us the initial condition:

p(0) = 1000

Using the given information that when $100 is spent on advertising, the profit is $6,000 (p = 6000), we can find the value of k:

dp/dx = k(11000 - p) = k(11000 - 6000) = 5000k = (dp/dx) when x = 100

Since dp/dx = (p - 1000) / 100 and (p - 1000) / 100 = 5000k, we can solve for k:

(p - 1000) / 100 = 5000k

p - 1000 = 500000k

6000 - 1000 = 500000k

5000 = 500000k

k = 5000 / 500000 = 0.01

Now, we can solve the initial-value problem to determine the profit if $200 were spent on advertising:

dp/dx = 0.01(11000 - p)

Separating variables and integrating:

1 / (11000 - p) dp = 0.01 dx

Integrating both sides:

ln|11000 - p| = 0.01x + C

Now, we can use the initial condition p(0) = 1000:

ln|11000 - 1000| = 0.01(0) + C

ln|10000| = C

C = ln(10000)

Substituting the value of C back into the equation:

ln|11000 - p| = 0.01x + ln(10000)

Taking the exponential of both sides:

e^(ln|11000 - p|) = e^(0.01x + ln(10000))

11000 - p = e^(0.01x) * 10000

11000 - p = 10000 * e^(0.01x)

Solving for p, when x = 200:

p = 11000 - 10000 * e^(0.01 * 200)

Calculating this value gives us the profit if $200 were spent on advertising.