I saw the same type of questions are asked for that but I am new at this subject so I couldn't grasp it totally.
so basically for that production function:
$$ f(x_1,x_2) = min \{ 2x_1 +x_2 , x_1 +2x_2 \} $$ what is the way to find $C(w,y)$ ?
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Sign up to join this communityCost minimisation problem of the firm is defined as follows:
\begin{eqnarray*} \min_{x_1\geq 0, x_2\geq 0} & w_1x_1+w_2x_2 \\ \text{s.t. } & \min(2x_1+x_2, x_1+2x_2) = y \end{eqnarray*}
where $w_1>0$, $w_2>0$, $y>0$ are given.
Here is the graph of the constraint set:
Observe that the solution to the above problem is given by: \begin{eqnarray*} (x_1^c, x_2^c) \in \begin{cases} \left\{(y,0)\right\} & \text{if } \frac{w_1}{w_2} < \frac{1}{2} \\ \left\{t(y,0)+(1-t)\left(\frac{y}{3},\frac{y}{3}\right)|0\leq t\leq 1\right\} & \text{if } \frac{w_1}{w_2} = \frac{1}{2} \\ \left\{\left(\frac{y}{3},\frac{y}{3}\right)\right\} & \text{if } \frac{1}{2} < \frac{w_1}{w_2} < 2 \\ \left\{t(0,y)+(1-t)\left(\frac{y}{3},\frac{y}{3}\right)|0\leq t\leq 1\right\} & \text{if } \frac{w_1}{w_2} = 2 \\ \left\{(0,y)\right\} & \text{if } \frac{w_1}{w_2} > 2\end{cases} \end{eqnarray*}
Consequently, the cost function (which is the optimal cost) is \begin{eqnarray*} C(w_1,w_2,y) = \begin{cases} w_1y & \text{if } \frac{w_1}{w_2} \leq \frac{1}{2} \\ (w_1+w_2)\frac{y}{3} & \text{if } \frac{1}{2} < \frac{w_1}{w_2} < 2 \\ w_2y & \text{if } \frac{w_1}{w_2} \geq 2\end{cases} \end{eqnarray*}