Thread: teach me do this question pls. I think this question about economies and math

1. teach me do this question pls. I think this question about economies and math

A firm has using two inputs to produce a single output, Q . The Firm’s production function is Q = x11/2 x21/2 If input price of x1 and x2 are MYR2 and MYR8 per unit, respectively, get the necessary conditions and use it to determine :

(a) Quantity of x1 and x​2 that minimizes the cost
(b) Minimum cost if the firm produces output of 20 units

2. Re: teach me do this question pls. I think this question about economies and math

Originally Posted by dDmondChen
A firm has using two inputs to produce a single output, Q . The Firm’s production function is Q = x11/2 x21/2 If input price of x1 and x2 are MYR2 and MYR8 per unit, respectively, get the necessary conditions and use it to determine :

(a) Quantity of x1 and x​2 that minimizes the cost
(b) Minimum cost if the firm produces output of 20 units
What is the cost function? Is it $2x_1+8x_2$?

3. Re: teach me do this question pls. I think this question about economies and math

why need to find cost function? im using MPL/w=MPK​/r

4. Re: teach me do this question pls. I think this question about economies and math

Originally Posted by dDmondChen
why need to find cost function? im using MPL/w=MPK​/r
Those symbols may be used in your textbook, but they are not familiar to me. I will assume that the "cost" is as I described.

(a) Then, the cost is obviously minimized when the quantity produced is 0 and $x_1=x_2=0$.

(b) You want to minimize $2x_1+8x_2$ given the conditions that $\sqrt{x_1x_2} = 20$. Solve for $x_2$. $x_2 = \dfrac{400}{x_1}$. Plug that in to the cost function and take the derivative with respect to $x_1$. Set the derivative equal to zero to find critical points:

$2-\dfrac{3200}{x_1^2} = 0$

$x_1 = 40, x_2=10$

Next, we need to show this yields the minimum cost. The instantaneous rate of change in cost at $x_1=20$ is $2-\dfrac{3200}{400} = -6<0$ and at $x_1=80$, it is $2-\dfrac{3200}{6400} = \dfrac{3}{2}>0$. This shows that a minimum occurs at $x_1=40$, and the cost is minimized at the values we found.