Ejercicio #1: Equilibrio Isocuanta e Isocosto - YouTube

Okey let´s do an equilibrium exercise of isoquants

and isocosts

which says the following... let's go to the exercise

what does the exercise offer us

the balance exercise ask the

product maximization subject to a cost

which means the exercise will plant a given cost

and we have to find the function production

equivalent to isoquant

that maximizes the production

given the cost

so it gives us a production function

Q = -L^3 + 6KL^2

and it gives us the total cost is 1000 dollars

the worker salary

ten dollars

the capital cost R is 20 dollars

with this information we have to find the K value

the L value

that maximizes the production

according to the producer theory

the long-term production function

the isoquants production

the isocosts curve

that is tangent to isoquants

is the one that maximizes production

let's graphic the rent of isocosts

let's find the first value

the first value means we are gonna asign

all the resources of one thousand dollars

we are not gonna adquire L but only K

for this we know the total cost / capital cost

replace

1000 / 20

we have that point is equal to 50 K

let's find isocosts

I asign all the resources to 0 K

and all the resources I asign it to L

we know the formula total cost

divided by work cost which is 'w'

1000 / 10

is 100

If I join the two points we have the isocosts curve

we have to find the point where the isocosts point

is tangent to the isoquants curve

in order to produce we have to find

the 'L' value

'K' value of that point

and find the production

given the 'K' value

we have graph

let's proceed then to find the values

that help me calculate

the values of K and L

we see in the ecuation that 'L' marginal productivity

divided by 'K' marginal productivity

is equal to cost of work

divided by cost of capital

so do you remember the theory

the worker marginal productivity

we are gonna obtain it with the partial derivative

in what consists, we are gonna start with the original production function

Q=-L^3+6KL^2

when I'm going to work with only one derivative partial

only derivate the terms

that contain L

which means marginal productivity

is equal to -L^3+6KL^2

which is equal to -3L^3-1 + 12KL^2-1

solving this I got -3KL^2+12KL

So now we got the ecuation that represents the marginal productivity of 'L'

which is euqla to -3L^2 + 12KL

now we are going to obtain the K marginal productivity

which is equal to partial derivative

we start from the production original equation

Q=-L^3+6KL^2

and we are going to derive the terms that only have K

because we are gonna calculate the K marginal productivity

so just the first term does not have K

which means he does not take it into consideration

I'm only going to derive the term 6KL^2

which K is raised to 1

and we have 6K^1-1L^2

remember 6L^2 we leave there

we have 6K^1-1 is 0 which is 1

and we have 6(1)L^2 equals to 6L^2

and in that way we have an equation that represents

K marginal productivity

that is equal to 6L^2

having the two equations

of marginal productivity of K

we are gonna replace the values

of w and r

we have that the productivity of L is 3L^2+12KL

divided by 6L^2

twhich are the values obtained previously

is equal to salary cost 10 and capital cost r

the 20 dividing pass to multiply

to the term -3L^2+12KL

6L^2 passes to multiply the 10

solving this we have -60L^2+240KL=6L^2(10)

we are gonna pass the term .... we have -60L^2 and 60L^2

which is -120L + 240KL

equal to 0

L(-120L+240K)=0

and check this out

If L multiplies this term, is equal to 0

it means that term is equal to 0

L*0 equal to 0

that term -120L+240K we are gonna take it out

and equal it to 0

Let's clear L

L=-240K/-120

L=2K

we are gonna replace this value on the equation that the total cost

is equal to the capital

plus the worker cost

multiplied by the quantity worked

how much is the total cost, is going to be equal to 1000=20*K+10*2K

1000=20K+20K

we have 40K=1000

K=1000/40

K=25

we have our maximum level of production

we have a value, K is equal to 25

what I do to replace L

I simply replace it in the equation

L=2K

L=2(25)

L=50

So I have that when L is equal to 50 and K equal to 25 I obtain the maximum level of production

I'm gonna replace this values

in the original function

Q=-L^3+6KL^2

K=25

L=50

Replace L and K

I have Q=-(50)^3+6(50)25^2

solving that I have Q=-125000+375000

units. The maximum production we could have

When K=25 and L=50

is 250,000 units

so we have not only obtain a value to find that point

where the isocosts curve is tangent to the isoquant curve

Where K=25 and L=50

in that way we have conclude

the exercise

solving the equilibrium, maximizing the production

given a cost