<p>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} &amp; P(a - g(u)) + (1-P)(b-g(v)) \\ \text{s.t.} &amp; 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} &amp; Pg(u) + (1-P)g(v) \\ \text{s.t.} &amp; 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} } &amp; Pg(u) + (1-P)g\left(\frac{c - Pu}{1-P}\right) \end{eqnarray*} 
Differentiating with respect to $u$, we get the FOC as:</p>

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

<p>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.</p>

CC BY-SA 3.0
2017-04-11T19:22:04.720Z
This is an excellent answer. Very helpful, thanks!
jlol
CC BY-SA 3.0
2017-04-11T10:22:44.070Z
2017-04-11T12:12:02.813Z
2017-04-11T12:12:02.813Z
Amit
<p>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*}</p>

<p>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*}</p>

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

<p>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.</p>

CC BY-SA 3.0
2017-04-11T19:23:08.007Z
This was of great help. Now I understand, thanks!
jlol
CC BY-SA 3.0
2017-04-11T20:46:50.530Z
@jlol my pleasure, I'm glad it was helpful.
Oliv
CC BY-SA 3.0
2017-04-11T10:12:25.833Z
2017-04-11T10:22:18.103Z
2017-04-11T10:22:18.103Z
Oliv
<p>I am trying to understand how the first order conditions for an interior solution of a maximization problem were derived using the <em>substitution</em> method. </p>

<p>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: $a,b,c&gt;0$, $P\in (0,1)$, $f:[0,+\infty]\to[0,+\infty]$, increasing and strictly concave over its domain.</p>

<p>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!</p>

<microeconomics><optimization>
Solving a maximization problem by substitution when the constraint is in implicit form
jlol
# Solving a maximization problem by substitution when the constraint is in implicit form

(Asked by: username_0 on 2017-04-11)

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!

**Tags:** microeconomics, optimization

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## 2 answers

### Answer by username_2 on 2017-04-11. Score: 4 | Accepted answer

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.

> **Comments:**
> **username_0**: This was of great help. Now I understand, thanks!  
> **username_2**: @jlol my pleasure, I'm glad it was helpful.  

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### Answer by username_1 on 2017-04-11. Score: 4

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.

> **Comments:**
> **username_0**: This is an excellent answer. Very helpful, thanks!  

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