I am struggling a bit with a Kuhn-Tucker maximization problem.
The problem is as follows:
$\max_{v} u(y_{h} + vy_{w}) - g(v)$
$\text{s. t. } u(y_{w}(1-v)) - h(v) \geq \overline{u}(y_{w})$
u() is increasing, concave, and h and g increasing convex. The outcome of interest is $\frac{dv}{dy_{w}}$ and $\frac{dv}{dy_{h}}$. It might be easier with an example, so let u(x) = log(x), and h(v)=g(v)=v^2. The Lagrange is then:
$L = \ln(y_{h} + vy_{w}) - v^2 + \lambda[ \ln(y_{w}(1-v)) - v^2 - \overline{u}(y_{w})]$
I've set up the Kuhn Tucker conditions:
(1) $\frac{y_{w}}{y_{h}+vy_{w}} - 2v + \lambda\left[\frac{-y_{w}}{y_{w}(1-v)} - 2v\right]=0 $
(2) $\lambda \geq 0$
(3) $\lambda\left[\ln(y_{w}(1-v)) - v^2 - \overline{u}(y_{w})\right] = 0$
(4) $\ln(y_{w}(1-v)) - v^2 \geq \overline{u}(y_{w})$
First, if $\ln(y_{w}(1-v)) - h(v) > \overline{u}(y_{w})$, we must have $\lambda=0$ so that condition (3) holds. Then condition (1) reduces to $\frac{y_{w}}{y_{h}+vy_{w}} - 2v = 0$, and I can easily solve for v, and consequently $\frac{dv}{dy_{w}}$ and $\frac{dv}{dy_{h}}$.
My problem arises when the constraint binds My questions are then as follows:
- Can I just total differentiate the constraint (in (4)) to find $\frac{dv}{dy_{w}}$, or do I need to consider condition (1)?
- If I need to total differentiate the condition (1), is the following correct:
$ \frac{y_{h}}{(y_{h} + vy_{w})^{2}}dy_{w} - \frac{y_{w}}{(y_{h} + vy_{w})^{2}}dy_{h} + \left[2\lambda -2 -\frac{\lambda}{(1-v)^{2}}\right]dv= 0 $
And thus, if dy_w = 0:
$\frac{dv}{dy_{h}} = \frac{\frac{y_{w}}{(y_{h} + vy_{w})^{2}}}{2\lambda - 2 - \frac{\lambda}{(1-v)^{2}}} $
and when dy_w = 0:
$\frac{dv}{dy_{w}} = \frac{\frac{y_{h}}{(y_{h} + vy_{w})^{2}}}{-2\lambda + 2 + \frac{\lambda}{(1-v)^{2}}}$
Thanks!
