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.

Answer.

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.