# Total output function

The total output (Yp) is a function of the amount of Labor (L), Capital (K), and Technology (A) you have available.

Let's apply this with the following scenario:

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

Yp = A * L½ * K½

You’ll notice that in the problem, it defines A for us at the end.

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

Therefore, we can go ahead and plug it into our equation like so:

Yp = 15 * L½ * K½

For context, this value represents changes in technology in our dog factory. For example, if we upgraded our machine, our A value might be higher at 20. If we had worse machines, it might be lower at 10.

You'll never have to calculate A, it’ll always be given to you!

Yp = 15 * L½ * K½

To find L, we’re going to zone in on the following part of the problem:

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

Essentially, we need to know where Labor Demand and Labor Supply intersect on a supply and demand graph.  The point of intersection is Wage Equilibrium (because Labor Demand and Labor Supply are determined by wages).

Utilizing this point of intersection, we can find where L falls on the x-axis.

Visually, we’ve identified on the graph where the Wage Equilibrium point occurs (and therefore where our L value lies on the x-axis).

But how can we solve for this value mathematically?

By setting the Labor Supply and Labor Demand equations…

Labor Demand: W = 100 – Ld
Labor Supply: W = 73 + 2Ls

…equal to each other...

100 – Ld = 73 + 2Ls

...and solving for L!

100 - L = 73 + 2L
27 = 3L
L = 9

In simple terms: By setting the wage functions equal we are essentially asking… “at equilibrium, what amount of labor do we have available for production?”. And when we solve for that, we get an L value of 9!

Let’s go ahead and plug that into our overarching production function like so:

Yp = 15 * 9½ * K½

Yp = 15 * 9½ * K½

To find K, we’re going to zone in on the following part of the problem:

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

To solve for K, we’re going to use the following formula:

K (this year) = K (last year) + Gross Investment - Depreciation

Considering this, all we need to do is plug in last year’s capital stock of 129…

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

K (this year) = 129 + Gross Investment - Depreciation

…the gross investments of 25…

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

K (this year) = 129 + 25 - Depreciation

…and the depreciation value of 10…

Question: You’ve determined the Labor Demand equation for your dog food factory is W = 100 – Ld, and the Labor Supply equation is W = 73 + 2Ls. What is the equilibrium value for L (Labor)? Additionally, Capital Stock from last year was 129, gross investments were 25, and depreciation was 10. What is the value for K (Capital)? Utilizing these variables, determine the total output (Yp), given that A is 15.

K (this year) = 129 + 25 - 10

…to result in a K value of 94!

K (this year) = 129 + 25 - 10 = 144

Let’s go ahead and plug that into our total output equation like so:

Yp = 15 * 9½ * 144½

Yp = 15 * 9½ * 144½

This results in a total output (Yp) of 540!

Yp = 15 * 9½ * 144½
Yp = 15 * 3 * 12
Yp = 540

This means that based on our amount of labor, capital, and technology available, we can product 540 units of dog food.

This concept does not only apply to small-scale production, it can also be applied to much larger scales—such as nations and countries. Using this same framework, in theory, we can determine the potential output of a large economy by utilizing a production function.