Profit Maximization Graphing

[faeralini]

I have an isoprofit curve equation that is as such:

[FONT=arial, sans-serif]max π = p * f(x[/FONT]₁[FONT=arial, sans-serif], x[/FONT]₂[FONT=arial, sans-serif]) - x[/FONT]₁[FONT=arial, sans-serif]w[/FONT]₁[FONT=arial, sans-serif] - x[/FONT]₂[FONT=arial, sans-serif]w[/FONT]₂
[FONT=arial, sans-serif][/FONT]
[FONT=arial, sans-serif]where[/FONT]
[FONT=arial, sans-serif]p = output price[/FONT]
[FONT=arial, sans-serif]x[/FONT]₁[FONT=arial, sans-serif] = input 1[/FONT]
[FONT=arial, sans-serif]x[/FONT]₂ [FONT=arial, sans-serif]= input 2[/FONT]
[FONT=arial, sans-serif]w[/FONT]₁[FONT=arial, sans-serif] = input 1 price[/FONT]
[FONT=arial, sans-serif]w[/FONT]₂[FONT=arial, sans-serif] = input 2 price[/FONT]

[FONT=arial, sans-serif]I believe total differentiation would go like this

0 = p * [ (df/dx₁) * dx₁ + (df/dx₂) * d₂] - w₁dx₁ - w₂dx₂

0 = (pf₁ - w₁)dx₁ + (pf₂ - w₂)dx₂

dx₂/dx₁ = - [ (pf₁ - w₁) / (pf₂ - w₂) ]

Now I want to show this in excel as a table and a graph (i believe it would show up as circular contour lines?) where i can enter a value for x₁ and it will tell me what x₂ should be for profit maximization

It doesn't matter what the constants are so any example would work

If I said
f(x₁, x₂) = x₁^.4 x₂^.6
p = 6
w₁ = 1
w₂ = 3
[/FONT]π = 60

60 = 6 * x₁^.4x₂^.6 - x₁ - 2x₂

This is where I'm stuck. I'm having trouble isolating either x₁ or x₂ for use as an excel equation

Am I missing a step or something?

 

[JeffM]

I have an isoprofit curve equation that is as such:

max π = p * f(x₁, x₂) - x₁w₁ - x₂w₂

where
p = output price
x₁ = input 1
x₂ = input 2
w₁ = input 1 price
w₂ = input 2 price

I believe total differentiation would go like this

0 = p * [ (df/dx₁) * dx₁ + (df/dx₂) * d₂] - w₁dx₁ - w₂dx₂

0 = (pf₁ - w₁)dx₁ + (pf₂ - w₂)dx₂

dx₂/dx₁ = - [ (pf₁ - w₁) / (pf₂ - w₂) ]

Now I want to show this in excel as a table and a graph (i believe it would show up as circular contour lines?) where i can enter a value for x₁ and it will tell me what x₂ should be for profit maximization

It doesn't matter what the constants are so any example would work

If I said
f(x₁, x₂) = x₁^.4 x₂^.6
p = 6
w₁ = 1
w₂ = 3
π = 60

60 = 6 * x₁^.4x₂^.6 - x₁ - 2x₂

This is where I'm stuck. I'm having trouble isolating either x₁ or x₂ for use as an excel equation

Am I missing a step or something?

If I understand what you are asking, I am totally unable to tell you whether Excel can do contour mapping and if so how to do it.

Furthermore, I do not believe that Excel can do partial differentiation.

Finally, you have not specified boundary conditions, which may be necessary to find solutions.

Let's take your example \(\displaystyle f(x_1,\ x_2) = x_1^4 * x_2^6 \implies f_1(x_1,\ x_2) = 4x_1^3x_2^6\ and\ f_2(x_1,\ x_2) = 6x_1^4x_2^5.\)

\(\displaystyle \pi = p * f(x_1,\ x_2) - w_1x_1 - w_2x_2.\)

Provided the second partial derivative is nicely behaved, a local maximum will occur where:

\(\displaystyle 0 = 4px_1^3x_2^6 - w_1\ and\ 0 = 6px_1^4 x_2^5 - w_2 \implies x_1 > 0 \le x_2 = \sqrt[6]{\dfrac{w_1}{4px_1^3}}\ and\ x_2 = \sqrt[5]{\dfrac{w_2}{6px_1^4}}.\)

You have not specified boundary conditions so there will only be a local maximum if:

\(\displaystyle \sqrt[6]{\dfrac{w_1}{4px_1^3}} = \sqrt[5]{\dfrac{w_2}{6px_1^4}} \implies \dfrac{\sqrt[5]{x_1^4}}{\sqrt[6]{x_1^3}} = \dfrac{\sqrt[5]{w_2} * \sqrt[6]{4p}}{\sqrt[6]{w_1} * \sqrt[5]{6p}} \implies\)

\(\displaystyle \left(\dfrac{\sqrt[5]{x_1^4}}{\sqrt[6]{x_1^3}}\right)^{30} = \left(\dfrac{\sqrt[5]{w_2} * \sqrt[6]{4p}}{\sqrt[6]{w_1} * \sqrt[5]{6p}}\right)^{30} \implies \dfrac{x_1^{24}}{x_1^{15}} = x_1^9 = \left(\dfrac{\sqrt[5]{w_2} * \sqrt[6]{4p}}{\sqrt[6]{w_1} * \sqrt[5]{6p}}\right)^{30} \implies x_1^3 = \left(\dfrac{\sqrt[5]{w_2} * \sqrt[6]{4p}}{\sqrt[6]{w_1} * \sqrt[5]{6p}}\right)^{10} \implies\)

\(\displaystyle x_1 = \sqrt[3]{\left(\dfrac{\sqrt[5]{w_2} * \sqrt[6]{4p}}{\sqrt[6]{w_1} * \sqrt[5]{6p}}\right)^{10}}.\)
I think that is right, but I am not going to check it because I am fairly confident that with a completely open-ended set of functions, particularly ones this bizarre, excel is going to give up the ghost.

You need to limit the functions, say to a Cobb-Douglas production function where you can figure out the partial derivatives and solve everything using parameters.