Question

Amos Jones and Andrew Brown own and operate Amos & Andy, Inc., a Minneapolis-based installer of conversion packages for vans manufactured by the major auto companies. Amos & Andy has fixed capital and labor expenses of $1.2 million per year, and variable materials expenses average $2,000 per van conversion. Recent operating experience suggests the following annual demand relation for Amos & Andy products:

Q = 1,000 - 0.1P
where Q is the number of van conversions (output) and P is price.

A) Calculate Amos & Andy’s profit-maximizing output, price, and profit levels.

B) Using the Lagrangian multiplier method, calculate profit-maximizing output, price, and profit levels in light of a parts shortage that limits Amos & Andy’s output to 300 conversions during the coming year.

C) Calculate and interpret , the Lagrangian multiplier.

D) Calculate the value to Amos & Andy of having the parts shortage eliminated.

Answer 1

A.

Since Q = 1,000 - 0.1P, => P=$10,000 - $10*Q, => TR = PQ=$10,000*Q - $10Q^2

Now, here the fixed expenses “$1,200,000” per year and average variable costs “$2,000” per unit, the total cost function for the coming year is given by, “TC = $1,200,000 + $2,000*Q”.

The Amos & Andy profit function is given below.

=> π = TR – TC = $10,000*Q - $10Q^2- ($1,200,000 + $2,000*Q)

Now, at the optimum, dπ/dQ must be equal to “0”.

=> dπ/dQ = (-20)Q + 8,000 = 0, => 20Q= 8,000, => Q = 400.

So, at the optimum “Q” the corresponding “P” is given by, P = $10,000 - $10*Q

=> $10,000 - $10*(400) = $6,000.

Now, the corresponding “profit” is given below.

=> π= (-$10)*(400^2) + $8,000(400) - $1,200,000, = $400,000

B.

Now, with Amos & Andy output limited to Q = 300, the constraint is given by “0 = 300 – Q”, the optimization problem or the “lagrangian function” can then be written as follows.

=> π= -$10Q2+ $8,000Q - $1,200,000 +λ*(300 - Q), the FOC are given below.

=> ∂Lπ/Q = -20Q + 8,000 –λ = 0…………………(1)

=> ∂Lπ/∂λ = 300 - Q = 0………………..(2)

Now, solving these two equation simultaneously we will get the optimum “Q” and “λ”. So, here the optimum value of “Q = 300” and “λ=2000”.

So, the corresponding “P” of this optimum “Q” is given below.

=>P = $10,000 - $10*Q = $10,000 - $10(300) = $7,000, => P = $7,000.

Now, the optimum profit is given by,

=> π= -$10Q^2+ $8,000*Q - $1,200,000 = (-$10)*(300^2) + $8,000*(300) - $1,200,000 = $300,000.

So, the optimum profit is given by, “300,000”.

C.

We can see form the lagrange function that, λ = ∂π/∂Q = 2,000, which implied that profits would increase by “$2,000” if output were to increase by one unit.

D.

Now, with no vehicle shortage (in A) Amos & Andy earned $400,000, but only $300,000 with a shortage (in B).Thus, having the vehicle shortage eliminated has a maximum value of $100,000 (= $400,000 - $300,000) to the company.