Question

The quantity of a product demanded by consumers is a function of its price. The quantity of one product demanded may also depend on the price of other products. For example, if the only chocolate shop in town (a monopoly) sells milk and dark chocolates, the price it sets for each affects the demand of the other. The quantities demanded, q1​ and q2​, of two products depend on their prices, p1 and p2, as follows:
q1​=150−2p1​−p2
q2​=200−p1​−3p2​.​

If one manufacturer sells both products, how should the prices be set to generate the maximum possible revenue? What is that maximum possible revenue?

Answer 1

Answer:

Both prices should be set to #25.

The maximum revenue is #4375

Step-by-step explanation:

Given

$q_1 = 150-2p_1-p_2$

$q_2 = 200-p_1-3p_2$

Start by calculating the total revenue (R):

$R = p_1q_1 + p_2q_2$

$R = p_1(150-2p_1-p_2) + p_2(200-p_1-3p_2)$

$R = 150p_1-2p_1^2-p_1p_2 + 200p_2-p_1p_2-3p_2^2$

Collect and solve like terms

$R = 150p_1+ 200p_2-2p_1^2-2p_1p_2 -3p_2^2$

Differentiate with respect to pi and to p2, respectively

$\frac{dR}{dp_1} = 150 -4p_1 - 2p_2$

$\frac{dR}{dp_2} = 200 -2p_1 - 6p_2$

Equate both to 0, to get the critical point

$150 -4p_1 - 2p_2 = 0$

$200 -2p_1 - 6p_2 = 0$

Solve for p1 in $150 -4p_1 - 2p_2 = 0$

$4p_1 = 150 - 2p_2$

$p_1 = 37.5 - 0.5p_2$

Substitute $p_1 = 37.5 - 0.5p_2$ in $200 -2p_1 - 6p_2 = 0$

$200 - 2(37.5 - 0.5p_2) - 6p_2 = 0$

$200 - 75 - p_2 - 6p_2 = 0$

$125 - 7p_2 = 0$

$-7p_2 =-125$

$p_2 = 25$

Substitute $p_2 = 25$ in $p_1 = 37.5 - 0.5p_2$

$p_1 = 37.5 - 0.5 * 25$

$p_1 = 37.5 - 12.5$

$p_1 = 25$

So, we have:

$p_1 = p_2 = 25$

This implies that the prices should be set to #25.

The maximum possible revenue is:

$R = 150p_1+ 200p_2-2p_1^2-2p_1p_2 -3p_2^2$

$R = 150 * 25 + 200 * 25 -2 * 25^2 - 2 * 25 * 25 - 3 * 25^2$

$R = 4375$

The maximum revenue is #4375