I am trying to understand how the first order conditions for an interior solution of a maximization problem were derived using the substitution method.

The problem is: $$\max\limits_{x\ge0,y\ge0}P(a-x)+(1-P)(b-y)$$ subject to $$Pf(x)+(1-P)f(y)=c$$ where: $a,b,c>0$, $P\in (0,1)$, $f:[0,+\infty]\to[0,+\infty]$, increasing and strictly concave over its domain.

I can see how this is solved using a lagrangian to find from the first order conditions that $f'(x^*)=f'(y^*)$. Strict concavity of $f$ then implies $x^*=y^*$. But I'm at a loss as to how we can solve it by substituting the constraint in the objective function. Since $f$ is invertible, if $y$ did not appear in the constraint I would find $x$ from the constraint by inverting $f$ and substitute it in the objective function. Doing this here leads to complications that seem unnecessary for this simple problem. There has to be a simpler way that I cannot figure out: what is it? Thanks!


2 Answers 2


Here are two methods. First method: the substitution can be made by inverting $f$. Since $f$ is strictly increasing and continuous, $f^{-1}$ is well-defined. The constraint can therefore be written \begin{equation*} x = f^{-1}\Big(\dfrac{c-(1-P)f(y)}{P}\Big) \end{equation*}

The objective becomes \begin{equation*} \max_{y \geq 0}{P \Big[a-f^{-1}\Big(\dfrac{c-(1-P)f(y)}{P}\Big)\Big]+(1-P) (b-y)} \end{equation*}

The derivative of this expression with respect to $y$ equals \begin{align*} & (1-P) f'(y) (f^{-1})^{'}(\dfrac{c-(1-P)f(y)}{P})-(1-P) \\ = & (1-P) \dfrac{f'(y)}{f^{'} \circ f^{-1}(\dfrac{c-(1-P)f(y)}{P})}-(1-P) \\ & = (1-P) (\dfrac{f'(y)}{f'(x)}-1) \end{align*} And thus $f'(y^{*})=f'(x^{*})$ at the optimum.

Second method: since $f$ is invertible, we can do a change in variables and define $w=f(x)$ and $z=f(y)$. The problem then becomes \begin{equation*} \max_{w \geq 0, z\geq 0}{P(a-f^{-1}(w))+(1-P)(b-f^{-1}(z))} \end{equation*} subject to \begin{equation*} P w +(1-P)z=c \end{equation*} Substituing $w=(c-(1-P)z)/P$ in the problem delivers \begin{equation*} \max_{z\geq 0}{P(a-f^{-1}(\dfrac{c-(1-P)z}{P})+(1-P)(b-f^{-1}(z))} \end{equation*} Differentiating with respect to $z$ yields the same solution.

  • $\begingroup$ This was of great help. Now I understand, thanks! $\endgroup$
    – jlol
    Apr 11, 2017 at 19:23
  • $\begingroup$ @jlol my pleasure, I'm glad it was helpful. $\endgroup$
    – Oliv
    Apr 11, 2017 at 20:46

Let $g$ be the inverse function of $f$ defined over range of $f$. Notice that $g$ is increasing and strictly convex. We can rewrite the maximization problem as: \begin{eqnarray*} \max\limits_{u\geq 0, \ v \geq 0} & P(a - g(u)) + (1-P)(b-g(v)) \\ \text{s.t.} & Pu + (1-P)v = c\end{eqnarray*} where $u=f(x)$ and $v = f(y)$. Solving above is equivalent to solving \begin{eqnarray*} \min\limits_{u\geq 0, \ v \geq 0} & Pg(u) + (1-P)g(v) \\ \text{s.t.} & Pu + (1-P)v = c\end{eqnarray*} Now you may substitute $v= \displaystyle\frac{c - Pu}{1-P}$ and rewrite the problem as: \begin{eqnarray*} \min\limits_{0\leq u \leq \frac{c}{P} } & Pg(u) + (1-P)g\left(\frac{c - Pu}{1-P}\right) \end{eqnarray*} Differentiating with respect to $u$, we get the FOC as:

$\displaystyle Pg'(u) - Pg'\left(\frac{c - Pu}{1-P}\right) = 0$

Since $g$ is strictly convex, solution is: $u = \displaystyle\frac{c - Pu}{1-P}$ i.e. $u = v = c$. Therefore, at the optimum $x = y$ holds.

  • $\begingroup$ This is an excellent answer. Very helpful, thanks! $\endgroup$
    – jlol
    Apr 11, 2017 at 19:22

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