Question

The total benefits of the COMMIT program which is designed to encourage school children to commit themselves to a life free of drug addiction has been estimated to be B = 800q - 4q2, where q is the number of days the program is run in local schools. The total cost of running the program is $400 per day.

a. What is the maximum number of days that the COMMIT program can be run before net benefits turn negative?

b. What is the optimal number of days to run the program in order to maximize net benefits? (MB = 800 - 8q.)

c. At the optimal number of days, what is the net benefit of the program?

d. If the cost of running the program were to rise to 500 per day, how would that affect the optimal number of days the program should be run?

Answer 1

a. Total benefit B = 800q - 4q² (where q is number of days program is run in local schools)

The maximum number of days that COMMIT program can be run before net benefits turn negative is B=0

or. 800q - 4q² = 0

or, 4q² = 800q

or, q = 200 days

b. In order to maximize net benefits, marginal benefit = 0

or, dB/dq = 0

or, 800-8q = 0

or, q = 800/8

or, q = 100 is the optimal number of days.

c. At the optimal number of days, net benefit of the program = 800q - 4q² = 800*100 - 4*100² = 80,000-40,000 = 40,000

d. Now, if cost of running the program were to rise to $500 per day,

net benefit = total cost

or, 800q-4q² = 500q

or, 4q² = 300q

or, q = 75 is the new optimal number of days

Answer 2

To find the answers to the questions, we'll analyze the total benefits (B), total costs (C), and net benefits (NB) of the COMMIT program. The net benefits are calculated as B - C.

Given: B = 800q - 4q^2 (Total benefits) C = 400q (Total cost)

a. To find the maximum number of days the COMMIT program can be run before net benefits turn negative, we need to determine the value of q when NB = 0 (i.e., B - C = 0).

NB = B - C = 800q - 4q^2 - 400q

Setting NB = 0:

800q - 4q^2 - 400q = 0

Solving for q:

4q^2 - 400q + 800q = 0

4q^2 + 400q = 0

4q(q + 100) = 0

q = 0 (Not considered in this context) or q = -100 (Negative, not considered)

Since q cannot be negative, the maximum number of days the program can be run before net benefits turn negative is 0 days.

b. To find the optimal number of days to run the program in order to maximize net benefits, we need to find the value of q that maximizes the net benefits. This corresponds to the vertex of the parabolic function NB = 800q - 4q^2.

To find the optimal q, we can use the formula for the x-coordinate of the vertex of a parabola, which is given by x = -b / 2a, where the quadratic equation is ax^2 + bx + c = 0.

For our equation NB = 800q - 4q^2:

a = -4, b = 800

q_optimal = -b / 2a = -800 / (2 * -4) = 100

So, the optimal number of days to run the program to maximize net benefits is 100 days.

c. To find the net benefit of the program at the optimal number of days (q_optimal = 100 days):

NB = 800 * 100 - 4 * 100^2 - 400 * 100 NB = 80000 - 4 * 10000 - 40000 NB = 80000 - 40000 - 40000 NB = 0

The net benefit of the program at the optimal number of days is 0.

d. If the cost of running the program were to rise to $500 per day, we need to redefine the total cost function (C_new).

Given: C_new = 500q (New total cost)

We want to find the new optimal number of days to run the program (q_new_optimal).

NB_new = B - C_new = 800q - 4q^2 - 500q

Setting NB_new = 0:

800q - 4q^2 - 500q = 0

4q^2 - 500q + 800q = 0

4q^2 + 300q = 0

4q(q + 75) = 0

q = 0 (Not considered in this context) or q = -75 (Negative, not considered)

Since q cannot be negative, the new optimal number of days to run the program with the increased cost (C_new = $500 per day) is 0 days.

In conclusion, with the increased cost of $500 per day, the optimal number of days to run the program is now 0 days. This means that it is not financially viable to run the program at all under the new cost condition.