# Thread: multivariable function maximisation and (?) Lagrange multiplier

1. ## multivariable function maximisation and (?) Lagrange multiplier

This is a sample exam question to which I don't have a model answer. I'd appreciate your feedback.

Question.

It is thought that a consumer measures the utility $\displaystyle u$ of posessing a quantity $\displaystyle x$ of apples and a quantity $\displaystyle y$ of oranges by the formula:

$\displaystyle u=u(x,y)=x^{\alpha}y^{1-\alpha}$.

It is know that, when a consumer's budget for apples and oranges is $1, he will buy 2 apples and 1 orange when they are equally priced. Find$\displaystyle \alpha$. The price of oranges falls to half that of apples with the price of apples unchanged. How many apples and oranges will the consumer buy for$10?

[Hint. First solve a utility maximisation problem with $\displaystyle \alpha$ as a parameter.

First part, finding alpha.

I will maximise $\displaystyle u(x,y)==x^{\alpha}y^{1-\alpha}$ by
first - find critical points by setting the first derivative to zero, then
second - evaluate the second derivative at this critical point(s)

$\displaystyle \nabla{u(x,y)}=({\frac{\partial{u}}{\partial{x}}}{ \frac{\partial{u}}{\partial{y}}})^T=(\alpha{x}^{\a lpha-1}y^{1-\alpha}, x^{\alpha}(1-\alpha)y^{-\alpha})^T=(0,0)^T$

$\displaystyle \alpha{x}^{\alpha-1}y^{1-\alpha}=x^{\alpha}(1-\alpha)y^{-\alpha}$

Then I use the fact that for $1 consumer would buy 2 apples and 1 orange, so I substitute x=2 and y=1:$\displaystyle \alpha{2}^{\alpha-1}1^{1-\alpha}=2^{\alpha}(1-\alpha)1^{-\alpha}\displaystyle \alpha2^{\alpha-1}=2^{\alpha}-2{\alpha}\alpha\displaystyle \alpha*2^{\alpha-1}-2^{\alpha}+2^{\alpha}*{\alpha}=0\displaystyle 2^{\alpha}(\frac{\alpha}{2}-1+\alpha)=0$Now when I try to evaluate the second derivative, it is Hessian matrix and it's determinant is 0. Therefore, the test is inconclusive. What can I do here to show that alpha=2/3 gives the maximum value of the function? (if it does...) Given$\displaystyle \alpha=2/2$and the utility function is$\displaystyle u=u(x,y)=x^{\frac{2}{3}}y^{1-\frac{2}{3}}$I will post the second part as a separate post, as this one is growing long. 2. Now that I have the alpha I will find x,y subject to the budgetary constraint of$10. I plan to use Lagrange multiplier method.

The price of apples is $1/3 as before, but the price of oranges is$1/6 (half of the price of apples).

Let budget function $\displaystyle g(x)=\frac{1}{3}x+\frac{1}{6}y=10$

Then I need to find a common tangent point of u(x,y) and g(x,y) subject to the constraint g(x,y)=10. Let $\displaystyle \lambda$ be the Langrange multiplier:

$\displaystyle \nabla{u}=\lambda\nabla{g}$

$\displaystyle (2/3x^{-1/3}y^{1/3}, 1/3x^{2/3}y^{-2/3})^T=\lambda(1/3, 1/6)^T$

$\displaystyle 2/3x^{-1/3}y^{1/3}=\lambda{1/3}$

$\displaystyle 1/3x^{2/3}y^{-2/3}=\lambda{1/6}$

After raising into 3rd power, I have $\displaystyle y/x=\lambda^3/8$ and I am not sure how to proceed next...

3. "solve" for lambda and form a relationship between x and y, then insert that into g.

$\displaystyle \lambda ={2y^{1/3}\over x^{1/3}} ={2x^{2/3}\over y^{2/3}}$

so y=x now insert that into g.

4. Indeed!

If x=y, 1/3x+1/6x=10 then x=20 and y=20. Solved!

5. you should have taken my calculus and probability classes
it would have been fun having you as a student
I just published another paper in Taiwan, my 18th at Academia Sinica

6. I wish I could! well i am grateful for the opportunity to learn from you here