Question

Your company manufactures two models of speakers, the Ultra Mini and the Big Stack. Demand for each depends partly on the price of the other. If one is expensive, then more people will buy the other. If p1 is the price of the Ultra Mini, and p2 is the price of the Big Stack, demand for the Ultra Mini is given by q1(p1, p2) = 100,000 − 200p1 + 10p2 where q1 represents the number of Ultra Minis that will be sold in a year. The demand for the Big Stack is given by q2(p1, p2) = 150,000 + 10p1 − 200p2. Find the prices for the Ultra Mini and the Big Stack that will maximize your total revenue. (Round your answers to two decimal places.) p1 = $ p2 = $

Answer 1

Answer:

p1 = $259.53   p2 = $381.20

Explanation:

1. Find the revenue function.

This is a typical income maximization problem. Therefore, the first thing we should know is what are the revenues for each product.

Recall that the revenue is given by P * Q

1.a Find the revenue of the Ultra Mini (product 1):

$R_{1} = P_{1} Q_{1}$

$R_{1} =P_{1} (100,000 - 200P_{1} + 10P_{2} )$

$R_{1} =100,000P_{1} -200P_{1} ^{2} +10P_{2}P_{1}$

1.b Find the revenue of the Big Stack (product 2):

$R_{2} = P_{2} Q_{2}$

$R_{2} =P_{2} (150,000 + 10P_{1} - 200P_{2} )$

$R_{2} = 150,000P_{1+2} +10P_{1}P_{2} -200P_{2}^{2}$

2. Find the marginal revenues.

The revenue function must be derived from the price.

For product 1, we derive from P1:

$MR_{1} = 100,000 -400P_{1} +10P_{2}$

For product 2, we derive from P2:

$MR_{2} = 150,000 + 10P_{1} - 400P_{2}$

3. Create a system of linear equations in two unknowns

With the marginal revenue functions we create a system of linear equations in two unknowns (p1 and p2) and equal 0.

$100,000 - 400P_{1} +10P_{2} = 0\\150,000 + 10P_{1} -400P_{2} = 0$

4. Resolve the previous system

4.a. To make it easier, we can rethink the terms of the system like this:

$100,000 - 400P_{1} +10P_{2} = 0$ is the same as saying:

$P_{2} = \frac{-100,000 + 400P_{1} }{10}$

And $150,000 + 10P_{1} -400P_{2} = 0$ is the same as saying:

$P_{2}=\frac{150,000+10P_{1} }{400}$

Therefore:

$\frac{-100,000 + 400P_{1} }{10} =\frac{150,000+10P_{1} }{400}$

Notice that now we only have one unknown (P1).

4.b. In order to eliminate fractionals, we can multiply both terms by 400:

$\frac{400}{10} (-100,000 + 400P_{1} ) = \frac{400}{400} (150,000 + 10P_{1} )$

$(40)(-100,000+400P_{1}) =150,000+10P_{1}$

$-4,000,000+16,000P_{1} =150,000+10P_{1}$

4.c. We solve the equation, putting numbers on one side and unknowns on the other:

$-4,000,000-150,000=10P_{1} -16,000P_{1}$

$-4,150,000=-15,990P_{1}$

$\frac{-4,150,000}{-15,990} =P_{1}$

$P_{1} = 259.53$

4.d. Once P1 has been identified, we replace it in any of the terms of the original system of equations (those established in 4.a).

$P_{2}= \frac{-100,000+400(259.53)}{10}$

$P_{2} = 381.20$