# Computing optimum efforts

Consider the following cost function:

$$c(e_1, e_2) = (\beta_1e_1 + \beta_2e_2)^2$$

The value function is:

$$v = v_0 - [l_1(1-e_1) + l_2(1-e_2)]$$

How do I compute the optimum efforts $e_1$ and $e_2$, with $e_{i} \in [0,1]$ ? I seem to be ending up in an inconsistency.

If I look at $v - c = v_0 - [l_1(1-e_1) + l_2(1-e_2)] - (\beta_1e_1 + \beta_2e_2)^2$, and then I take the partial derivative with respect to $e_1$ and $e_2$ respectively, I obtain:

$$\beta_1e_1^* + \beta_2e_2^* = \dfrac{l_1}{2\beta_1} = \dfrac{l_2}{2\beta_2}$$

What am I missing? Does the inconsistency simply mean that the optimal efforts are corner points?

You are right. The problem here is a corner solution.

Let us define the optimal combination $X^* = \beta_1e_1^* + \beta_2e_2^*$.

The first derivative gives you

$$X^* = \dfrac{l_1}{2\beta_1}$$

whereas the second gives you

$$X^* = \dfrac{l_2}{2\beta_2}$$

They can only be true "by chance". This is, the existence of an interior solution cannot be assured for every possible parameterisation allowed by the parameter space.

Actually, if you take a look at the two functions, this is clear. The cost function generate linear isoquants. You can check this here. This is because the cost function is equivalent to $c = X^2$ and $X$ is a linear combination of both $e_1$ and $e_2$.

Similarly, the value function also gives linear isoquants. This is because the function has the form $e_1 = a + be_2$.

Then, the most likely case is that the isoquants will join in a corner. The interior solution happens just by chance. Notice that when this is the case, any interior solution is optimal! The equilibrium set is then infinite.