<p>I do not know what invertibility of <span class="math-container">$U$</span> means as <span class="math-container">$U$</span> is a real valued function of two variables.</p>
<p>The following is true however:</p>
<ul>
<li>If <span class="math-container">$U$</span> is continuous, then Berge's extreme value theorem tells us that a solution exists. This solution can be a correspondence of <span class="math-container">$(p,y)$</span> (multivalued) but it will be upper-hemicontinuous.</li>
<li>If in addition <span class="math-container">$U$</span> is strictly concave, then for any <span class="math-container">$(p,y)$</span> there is a  solution must be unique. To see this, assume the optimal solution is not unique. Then take two of them. By strict (quasi) concavity, the convex combination of these two bundles gives strictly higher utility. As the budget is linear, it is also feasible. But this contradicted the optimality of the two initial bundles.</li>
<li>As such, if <span class="math-container">$U$</span> is continuous and strictly (quasi)-concave, one can write <span class="math-container">$q_1(y, p_1, p_2)$</span> and <span class="math-container">$q_2(y,p_1, p_2)$</span> as the unique demand functions. From upper hemi-contiuity in the first bullet point, it follows that both of them will be continuous.</li>
</ul>
<p>In general it is not possible to get an expression for <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> without imposing a functional form on <span class="math-container">$U$</span>.</p>

CC BY-SA 4.0
2024-09-09T15:58:37.933Z
2024-09-09T15:58:37.933Z
tdm
<p>Well you already have an answer in your question. Without any specific functional form for utility the the general solution is;</p>
<p><span class="math-container">$$q_i^*=q_i(p_1,p_2,y), \quad \text{for } i=1,2$$</span></p>
<p>Where <span class="math-container">$q_i$</span> are chosen such that the following conditions and budget constraint are satisfied;</p>
<p><span class="math-container">$$\frac{\partial U}{\partial q_i}-p_i \lambda = 0$$</span></p>

CC BY-SA 4.0
2024-09-08T04:13:24.360Z
2024-09-08T04:13:24.360Z
1muflon1
<p>Suppose we have a general maximization problem of the form:</p>
<p><span class="math-container">$$\max_{q_1,q_2} U(q_1,q_2) \text{ subject to } p_1 q_1 + p_2 q_2 = y$$</span></p>
<p>Suppose I allow <span class="math-container">$U$</span> to be concave, increasing and invertible. What does the solution look like? It must be something in terms of the inverted function <span class="math-container">$U$</span> I'm guessing, but I can't get the expression without a specific functional form.</p>
<p>I know that in general you get something like
<span class="math-container">$$ MRS = p_j / p_k $$</span>
but I'm looking for a characterization in terms of something like
<span class="math-container">$$ q_1 = g\Big(y,p_1,p_2\Big) $$</span></p>

CC BY-SA 4.0
2024-09-08T11:14:48.717Z
Dont you need stronger condition to ensure existence of the function g? what about perfect substitutes?
Jesper Hybel
CC BY-SA 4.0
2024-09-09T17:20:33.680Z
There is no continuous invertible function whose domain includes an open subset of $\mathbb{R}^2$ with values in $\mathbb{R}$.
Michael Greinecker
<utility><optimization>
Maximise Arbitrary Utility Subject to Budget Constraint
Debreu
# Maximise Arbitrary Utility Subject to Budget Constraint

(Asked by: username_0 on 2024-09-08)

Suppose we have a general maximization problem of the form:

$$\max\_{q\_1,q\_2} U(q\_1,q\_2) \text{ subject to } p\_1 q\_1 + p\_2 q\_2 = y$$

Suppose I allow $U$ to be concave, increasing and invertible. What does the solution look like? It must be something in terms of the inverted function $U$ I'm guessing, but I can't get the expression without a specific functional form.

I know that in general you get something like
$$ MRS = p\_j / p\_k $$
but I'm looking for a characterization in terms of something like
$$ q\_1 = g\Big(y,p\_1,p\_2\Big) $$

**Tags:** utility, optimization

> **Comments:**
> **username_1**: Dont you need stronger condition to ensure existence of the function g? what about perfect substitutes?  
> **username_2**: There is no continuous invertible function whose domain includes an open subset of $\mathbb{R}^2$ with values in $\mathbb{R}$.  

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## 2 answers

### Answer by username_3 on 2024-09-09. Score: 4

I do not know what invertibility of $U$ means as $U$ is a real valued function of two variables.

The following is true however:

* If $U$ is continuous, then Berge's extreme value theorem tells us that a solution exists. This solution can be a correspondence of $(p,y)$ (multivalued) but it will be upper-hemicontinuous.
* If in addition $U$ is strictly concave, then for any $(p,y)$ there is a solution must be unique. To see this, assume the optimal solution is not unique. Then take two of them. By strict (quasi) concavity, the convex combination of these two bundles gives strictly higher utility. As the budget is linear, it is also feasible. But this contradicted the optimality of the two initial bundles.
* As such, if $U$ is continuous and strictly (quasi)-concave, one can write $q\_1(y, p\_1, p\_2)$ and $q\_2(y,p\_1, p\_2)$ as the unique demand functions. From upper hemi-contiuity in the first bullet point, it follows that both of them will be continuous.

In general it is not possible to get an expression for $q\_1$ and $q\_2$ without imposing a functional form on $U$.


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### Answer by username_4 on 2024-09-08. Score: 1

Well you already have an answer in your question. Without any specific functional form for utility the the general solution is;

$$q\_i^\*=q\_i(p\_1,p\_2,y), \quad \text{for } i=1,2$$

Where $q\_i$ are chosen such that the following conditions and budget constraint are satisfied;

$$\frac{\partial U}{\partial q\_i}-p\_i \lambda = 0$$


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