## Maximize a Cobb Douglas Production Function Using Lagrange Multipliers

Suppose we’re given the following Cobb-Douglas production

function, P of L comma K, where L is the units of labor, K is units of capital, and P of L comma is total

units that can be produced with this labor/capital combination. Suppose each unit of labor costs $900, and each unit of capital costs $1,800. Also suppose a total of 1

million, 80 thousand dollars is available to be invested

in capital and labor combined. How many units of labor and

capital should be purchased to maximize production subject

to the budgetary constraint, and also what is the maximum

number of units of production under the given budgetary conditions? We want to maximize P of L comma K, under the given constraint. Let’s write the constraint as an equation. Again, L is the units of labor, and K is the units of capital. Labor costs $900 a unit, and capital costs $1,800 per unit, which means 900L plus 1,800K must equal the total budget of 1 million, 80 thousand dollars. Let’s go ahead and simplify

this by dividing through by 100. We can use the equivalent constraint, nine L plus 18K equals 10,800, and now we’ll use the method

of Lagrange Multipliers to maximize P under this constraint. In general, if you use the

method of Lagrange Multipliers, if we have a function f of x comma y, function f of x comma y is going to be maximized or minimized under the constraint g, when the gradient of f is

equal to some multiple, lamba times the gradient of g, which gives us these first two equations, and the third equation is our constraint, g of x comma y equals zero. In our case, the function f

is actually P of L comma K, and then for our

constraint, g of x comma y, we have g of L comma K, and because our constraint is

nine L plus 18K equals 10,800, to have our function equal to zero, we’ll subtract 10,800 on both sides, which would give us g of L comma K equals nine L plus 18K minus 10,800 equals zero, which means our system of

equations are going to be the partial of P with respect to L equals lambda times the

partial of g with respect to L, and we’ll have the partial

of P with respect to K equals lambda times the

partial of G with respect to K, and of course we’ll have our constraint, g of L comma K equals zero. Let’s go ahead and find

our partial derivatives. To find the partial of

P with respect to L, we’ll treat K as a constant, so we’re going to multiply by 0.9. 30 times 0.9 is 27, then we have L raised to

power of 0.9 minus one, that’s negative 0.1, times K to the 0.1 equals lambda times the

partial of g with respect to L, so we’ll treat K as a constant, so the partial with respect

to L would just be nine, so times nine. Now, for our second equation, we have the partial P with respect to K, so we differentiate

here with respect to K, treating L as a constant, so we’ll multiply by 1/10. 30 times 1/10 would be three L to the 0.9, K to the 0.1 minus one. That’d be negative 0.9 equals lambda times the

partial of g with respect to K, so we’re treating L as a constant, so the partial with respect

to K would just be eighteen. Of course G of L comma K equals zero is going to be nine L plus 18K minus 10,800 equals zero. Now we have to solve this

as a system of equations, so let’s go ahead and solve these first two equations for lambda. For the first equation, we

would divide both sides by nine, so we’d have lambda equals 27 L to the negative 0.1, K to the positive 0.1 divided by nine, and here we’d have lambda equals three L to the 0.9 K to negative 0.9 divided by 18. Now, because both of these

expressions are equal to lambda, we can set them equal

to each other and solve to determine the

relationship between L and K. When we do this, though, let’s

make the exponents positive, so we’ll move L to the negative

0.1 down to the denominator, as well as K to negative 0.9, so this would give us the equation 27 K to the 0.1 over, we still have a nine, and then we’d have L to

the positive 0.1 equals, here we’d have three L to the 0.9 divided by 18, and then we’d

have K to the positive 0.9. Again, this is true, because both of these are equal to lambda. Now we’ll cross multiply to determine the relationship between L and K, which means this product

must equal this product, so we’d have 27K to the 0.1 times 18K to the 0.9 must equal nine L to the 0.1 times three L to the 0.9. 27 times 18 is equal to 486, then K to the 0.1 times K to the 0.9 would be equal to just K. Remember, we’d be adding the exponents. 0.1 times 0.9 is equal to one, so we just have K to the first equals on the right,

nine times three is 27, and we have L to the

0.1 times L to the 0.9, which would be just L

to the first, or just L. To solve this equation for L, we’ll divide both sides by 27, which gives us L equals 486K divided by 27, would be equal to 18K. Now that we know that L is equal to 18K, what we’re going to do is substitute 18K for L in our constraint, which will give us one

equation and one variable, which will allow us to

find K as well as L. Let’s do this on the next slide. Again, we know L equals 18K, and we also know the constraint is nine L plus 18K minus 10,800 must equal zero, so if we substitute 18K for L, we have nine times 18K plus 18K equals 10,800. Well, nine times 18 is equal to 162, so we have 162K plus 18K, which equals 180K. Dividing both sides by 180, we have K equals 60. Well, now we know K equals

60 and we know L equals 18K, we know that L equals 18 times 60, which equals 1,080. Remember, L equals the units of labor, and K is equal to the units of capital. Going back to our question, we’re asked how many

units of labor and capital should be purchased to

maximize production, subject to the budgetary constraints. We know the units of

labor L is equal to 1,080, and the units of capital K is equal to 60. Now, to find the maximum

number of units of production under the given budgetary constraints, we need to substitute these values in for L and K into our function P. Let’s go ahead and do this. We’d have P of 1,080 comma 60, which would be equal to 30 times 1,080 raised to the 0.9 times 60 raised the power of 0.1. Going to the calculator,

we have 30 times 1,080 raised to the power of 0.9. Right arrow, open parenthesis 60, close parenthesis, raised to the power of 0.1. Rounding to the nearest unit, rounding down, we’d have 24,267, which means, under these constraints, the maximum production, again, is 24,267 units. Before we go, though, let’s take a look at what’s happening graphically. We’re going to graph the level curves for P of L comma K, as well as the constraint, g of L comma K. We see the level curves of P and g, graphed here in blue and green, and these two vectors here, the black vector and the red vector, are the gradient of P

and the gradient of g. Notice how where the level

curves are tangent to one another both the gradient of P

and the gradient of g are multiples of one another, which is the whole idea behind the method of Lagrange Multipliers, so this point here is the

point where the production is maximized under the given constraints. I hope you found this helpful.

Thank you!

how did you get A=30, Alpha =0.9? please help

You saved me. Best explanation ever.

Thanks

BY FAR the clearest explanation of this concept I've seen or read. Thanks so much!

So well done, thank you.

Thank you so much. By far the most useful video regarding Maximization using Lagrange multipliers.

Thanks a lot sir, can you also please explain how to Verify that these inputs are profit

maximising.

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Thanks!