<p>Objective function of Lagrangian can be set up either with <span class="math-container">$+\lambda$</span> or <span class="math-container">$-\lambda$</span>, depending on how you solve the budget constraint.</p>

<p>Actually, for the solution it does not matter if <span class="math-container">$\lambda$</span> has negative or positive sign in the equation. You can clearly see it from the formula if you expand the second term:
<span class="math-container">$$
\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p_{x}x - p_{y}{y})= U(x,y) +\lambda m -  \lambda p_{x}x - \lambda p_{y}{y}$$</span></p>

<p>Using the alternative approach you get exactly the same result as:</p>

<p><span class="math-container">$$
\mathcal{L}(x,y, \lambda) = U(x,y) - \lambda(p_{x}x + p_{y}{y}-m)= U(x,y) -  \lambda p_{x}x - \lambda p_{y}{y} +\lambda m$$</span></p>

<p>So the resulting equations are 100% equivalent.</p>

<p>However, it’s easy to confuse when to use which sign so many people just memorize one of the two ways of solving Lagrangian and never use the other, because it’s easier to just memorize one of the ways than every single time thinking about the sign of lambda. There is really nothing deeper about it.</p>

CC BY-SA 4.0
2020-01-13T23:47:32.423Z
2020-01-14T11:31:32.537Z
2020-01-14T11:31:32.537Z
1muflon1
<p>Lets say we have a utility function <span class="math-container">$U: \mathbb{R}^{2} \to \mathbb{R}$</span> given by <span class="math-container">$U(x,y)$</span> and a binding budget constraint <span class="math-container">$p_{x} x + p_{y} y = m$</span>, where <span class="math-container">$p_{x}, p_{y}$</span> are prices of goods <span class="math-container">$x,y$</span> and <span class="math-container">$m$</span> is income. We can maximize this function subject to our constraint by simply maximizing this the following function in the unconstrained sense: 
<span class="math-container">$$ 
\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p_{x}x - p_{y}{x})
$$</span>
where <span class="math-container">$\lambda$</span> is usually called the Lagrange Multiplier. </p>

<p>My question: The reason we're able to add the second term on the RHS of the expression above is because our budget constraint binds, i.e. <span class="math-container">$m - p_{x}x - p_{y}y = 0$</span>, so we're essentially adding zero. But, why is the conventional expression inside the parantheses after <span class="math-container">$\lambda$</span> given by <span class="math-container">$m - p_{x}x - p_{y}y$</span> and not <span class="math-container">$p_{x}x + p_{y}y - m$</span>? Shouldn't both expressions give us a zero term? My guess is that this has something to do with the sign of <span class="math-container">$\lambda$</span> (which we want to be positive because it can be interpreted as the marginal utility of <span class="math-container">$m$</span>), but I'm not sure how exactly this works, mathematically. Would appreciate a clarification explained mathematically. </p>

<microeconomics><utility><consumer-theory><optimization>
Budget Constraint in Utility Maximisation Problem with Lagrange Multipliers
gtoques
# Budget Constraint in Utility Maximisation Problem with Lagrange Multipliers

(Asked by: username_0 on 2020-01-13)

Lets say we have a utility function $U: \mathbb{R}^{2} \to \mathbb{R}$ given by $U(x,y)$ and a binding budget constraint $p\_{x} x + p\_{y} y = m$, where $p\_{x}, p\_{y}$ are prices of goods $x,y$ and $m$ is income. We can maximize this function subject to our constraint by simply maximizing this the following function in the unconstrained sense:
$$
\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p\_{x}x - p\_{y}{x})
$$
where $\lambda$ is usually called the Lagrange Multiplier.

My question: The reason we're able to add the second term on the RHS of the expression above is because our budget constraint binds, i.e. $m - p\_{x}x - p\_{y}y = 0$, so we're essentially adding zero. But, why is the conventional expression inside the parantheses after $\lambda$ given by $m - p\_{x}x - p\_{y}y$ and not $p\_{x}x + p\_{y}y - m$? Shouldn't both expressions give us a zero term? My guess is that this has something to do with the sign of $\lambda$ (which we want to be positive because it can be interpreted as the marginal utility of $m$), but I'm not sure how exactly this works, mathematically. Would appreciate a clarification explained mathematically.

**Tags:** microeconomics, utility, consumer-theory, optimization

---

## 1 answers

### Answer by username_1 on 2020-01-13. Score: 2

Objective function of Lagrangian can be set up either with $+\lambda$ or $-\lambda$, depending on how you solve the budget constraint.

Actually, for the solution it does not matter if $\lambda$ has negative or positive sign in the equation. You can clearly see it from the formula if you expand the second term:
$$
\mathcal{L}(x,y, \lambda) = U(x,y) + \lambda(m -p\_{x}x - p\_{y}{y})= U(x,y) +\lambda m - \lambda p\_{x}x - \lambda p\_{y}{y}$$

Using the alternative approach you get exactly the same result as:

$$
\mathcal{L}(x,y, \lambda) = U(x,y) - \lambda(p\_{x}x + p\_{y}{y}-m)= U(x,y) - \lambda p\_{x}x - \lambda p\_{y}{y} +\lambda m$$

So the resulting equations are 100% equivalent.

However, it’s easy to confuse when to use which sign so many people just memorize one of the two ways of solving Lagrangian and never use the other, because it’s easier to just memorize one of the ways than every single time thinking about the sign of lambda. There is really nothing deeper about it.


---

