Minimization of costs combination of factors of production

Question

Given the production function $D(x,y,z)=\min\{2x, \left( 2y+4z\right)\}$, and knowing that P-price $P_x=30,P_y=20,P_z=5$ and volume of production $D=200$. How much of x,y and z is needed to minimize the costs?

I know that i should write: $2x=2y+4z=200$ but how to find proper values for x,y and z? I general i think i should write this because those factors are complementary thus each of the "components" of the function should be equal to 200 which is the production volume.

The only thing i've managed to do is to guess the numbers: $x=100,z=50, y=0$. Is there a method to justify that those numbers will minimize the function?

Answer 1

Working with $\min$ (or $\max$) functions can be tricky, especially if one wants to write down the general solution in a fully rigorous way.

But in your case you have simple linear functions and a numerically given fixed production level ($200$) for which you want to minimize costs.

First, make sure you understand the production function $\min$ expression and what restrictions it poses: A frequently seen initial approach here is to think

"ah, this means either the one, or the other. So I will either use $x$ only, *or* a combination of $y$ and $z$ and perhaps only one from the two. There is no point in buying, say, $x$, if production is to be determined by the a combination of $y$ and $z$".

What is the fallacy in this line of reasoning? And what the resolution of the fallacy implies?

Next, think what it means to want to examine only a given fixed production level: on the right hand side you have the $\min$ function. On the left hand side, you have a specific number, not a general label/symbol anymore. So you have an equation, not a function anymore. And then you have the cost function. And you want to minimize the cost function given the equation, and given what the "fallacy resolution" mentioned previously implies.

Take it from there.

And please, once you figure it out, post it as an answer here (you can answer your own question is perfectly fine, and even encouraged).