<p>Check that the following utility function also represents the same preference as the one in the question:</p>
<p><span class="math-container">$u(x_1,x_2)=(x_1+1)(2x_2+1)$</span></p>
<p>Solving</p>
<p><span class="math-container">$\max_{(x_1,x_2)\in\mathbb{R}^2_+} (x_1+1)(2x_2+1)$</span> subject to <span class="math-container">$p_1x_1+p_2x_2\leq M$</span></p>
<p>we get</p>
<p><span class="math-container">\begin{eqnarray*} (x_1^d,x_2^d)(p_1,p_2,M)=\begin{cases}\left(0,\dfrac{M}{p_2}\right) &amp; \text{if } \dfrac{M}{p_2}+\dfrac{1}{2}&lt;\dfrac{p_1}{p_2} \\ \left(\dfrac{M}{p_1},0\right) &amp; \text{if } \dfrac{1}{2\left(\dfrac{M}{p_1}+1\right)}&gt;\dfrac{p_1}{p_2} \\ \left(\dfrac{M}{2p_1}-\dfrac{1}{2}+\dfrac{p_2}{4p_1},\dfrac{M}{2p_2}-\dfrac{1}{4}+\dfrac{p_1}{2p_2}\right)&amp; \text{if } \dfrac{1}{2\left(\dfrac{M}{p_1}+1\right)}\leq \dfrac{p_1}{p_2}\leq \dfrac{M}{p_2}+\dfrac{1}{2}\end{cases} \end{eqnarray*}</span></p>

CC BY-SA 4.0
2024-05-06T10:25:22.173Z
2024-05-06T10:25:22.173Z
Amit
<p>A consumer has the following utility function
<span class="math-container">$$u(x_1,x_2)=2x_1x_2+x_1+2x_2$$</span>
I want to maximize his utility function.
<span class="math-container">$$max: 2x_1x_2+x_1+2x_2.    uc:p_1x_1+p_2x_2=y_A$$</span>
Using Lagrange, I get
<span class="math-container">$$L(x_1,x_2,\lambda)=2x_1x_2+x_1+2x_2-\lambda(p_1x_1+p_2x_2-y_A)$$</span>
I then get the first order conditions:
<span class="math-container">$$\frac{d L}{d x_1}= 2x_2+1-\lambda p_1=0$$</span>
<span class="math-container">$$\frac{d L}{d x_2}= 2x_1+2-\lambda p_2=0$$</span>
<span class="math-container">$$\frac{d L}{d \lambda}= y_A-p_1x_1-p_2x_2=0$$</span></p>
<p>But I am not sure how to find the demand functions, ie. isolating <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>. I have tried dividing the first two terms with each other, but I am pretty much stuck at this point, and would greatly appreciate a hint in how to proceed</p>
<p>EDIT:
I have solved it thanks to a hint I received on this website, but I have one final question: My TA told me that it was important that I remember that this utility function is quasi-concave, but I don't understand why he would tell me that? Considering that my only task is to maximize utility and find the demand function.</p>

CC BY-SA 4.0
2024-05-06T07:12:51.267Z
There are three unknowns $(x_1, x_2, \lambda)$ and three equations so there are many ways to solve this. One way is to use the first two equations to eliminate $\lambda$. Then you have two equations left that only uses the variables $x_1$ and $x_2$.
tdm
CC BY-SA 4.0
2024-05-06T07:19:34.813Z
This is done by setting (1) and (2) equal to each other, right? And then solving for ?
Noah
CC BY-SA 4.0
2024-05-06T07:47:27.853Z
I have solved it, but I have one final question: My TA told me that it was important that I remember that the utility function is quasi-concave, but I don't understand why he would tell me that? Considering that my only task is to maximize utility and find the demand function.
Noah
CC BY-SA 4.0
2024-05-06T13:21:28.007Z
Quadi-concavity of the utility function (and linearity of the constraint) is a sufficient condition such that the first order conditions determine a global maximum. If the objective function is not quasi-concave, the solution that satisfies the first order conditions might not solve your utility maximization problem.
tdm
<microeconomics><mathematical-economics><optimization>
Solving utility maximization, and finding demand function
Noah
# Solving utility maximization, and finding demand function

(Asked by: username_0 on 2024-05-06)

A consumer has the following utility function
$$u(x\_1,x\_2)=2x\_1x\_2+x\_1+2x\_2$$
I want to maximize his utility function.
$$max: 2x\_1x\_2+x\_1+2x\_2. uc:p\_1x\_1+p\_2x\_2=y\_A$$
Using Lagrange, I get
$$L(x\_1,x\_2,\lambda)=2x\_1x\_2+x\_1+2x\_2-\lambda(p\_1x\_1+p\_2x\_2-y\_A)$$
I then get the first order conditions:
$$\frac{d L}{d x\_1}= 2x\_2+1-\lambda p\_1=0$$
$$\frac{d L}{d x\_2}= 2x\_1+2-\lambda p\_2=0$$
$$\frac{d L}{d \lambda}= y\_A-p\_1x\_1-p\_2x\_2=0$$

But I am not sure how to find the demand functions, ie. isolating $x\_1$ and $x\_2$. I have tried dividing the first two terms with each other, but I am pretty much stuck at this point, and would greatly appreciate a hint in how to proceed

EDIT:
I have solved it thanks to a hint I received on this website, but I have one final question: My TA told me that it was important that I remember that this utility function is quasi-concave, but I don't understand why he would tell me that? Considering that my only task is to maximize utility and find the demand function.

**Tags:** microeconomics, mathematical-economics, optimization
> **Comments:**
> **username_1**: There are three unknowns $(x_1, x_2, \lambda)$ and three equations so there are many ways to solve this. One way is to use the first two equations to eliminate $\lambda$. Then you have two equations left that only uses the variables $x_1$ and $x_2$.  
> **username_0**: This is done by setting (1) and (2) equal to each other, right? And then solving for ?  
> **username_0**: I have solved it, but I have one final question: My TA told me that it was important that I remember that the utility function is quasi-concave, but I don't understand why he would tell me that? Considering that my only task is to maximize utility and find the demand function.  
> **username_1**: Quadi-concavity of the utility function (and linearity of the constraint) is a sufficient condition such that the first order conditions determine a global maximum. If the objective function is not quasi-concave, the solution that satisfies the first order conditions might not solve your utility maximization problem.  

---

## 1 answers

### Answer by username_2 on 2024-05-06. Score: 2 | Accepted answer

Check that the following utility function also represents the same preference as the one in the question:

$u(x\_1,x\_2)=(x\_1+1)(2x\_2+1)$

Solving

$\max\_{(x\_1,x\_2)\in\mathbb{R}^2\_+} (x\_1+1)(2x\_2+1)$ subject to $p\_1x\_1+p\_2x\_2\leq M$

we get

\begin{eqnarray\*} (x\_1^d,x\_2^d)(p\_1,p\_2,M)=\begin{cases}\left(0,\dfrac{M}{p\_2}\right) & \text{if } \dfrac{M}{p\_2}+\dfrac{1}{2}<\dfrac{p\_1}{p\_2} \\ \left(\dfrac{M}{p\_1},0\right) & \text{if } \dfrac{1}{2\left(\dfrac{M}{p\_1}+1\right)}>\dfrac{p\_1}{p\_2} \\ \left(\dfrac{M}{2p\_1}-\dfrac{1}{2}+\dfrac{p\_2}{4p\_1},\dfrac{M}{2p\_2}-\dfrac{1}{4}+\dfrac{p\_1}{2p\_2}\right)& \text{if } \dfrac{1}{2\left(\dfrac{M}{p\_1}+1\right)}\leq \dfrac{p\_1}{p\_2}\leq \dfrac{M}{p\_2}+\dfrac{1}{2}\end{cases} \end{eqnarray\*}


---

