Need help with in optimisation

[j4m]

Struggling to get the first order conditions and would appreciate any help with the process. Cheers.

and budget constraint:

Here are the answers

 

[JeffM]

It is hard to be sure that I am doing this correctly when I have no idea what any of the variables represent. In fact, you have not even bothered to tell us what are variables and what are constants. Moreover, the utility equation has a very strange symbol in it. How many variables are we dealing with? Is BC an abbreviation for Before Christ? We are not mind readers.

It looks like a utility maximization problem subject to a budget constraint.

Consider the problem: maximize u subject to the constraint that $c_1 - c_2/R - y_1 - y_2/R = 0$, where all variables are positive and y_1, y_2, and R are constants.

$$L(c_1, \ c_2) = \ln(c_1) + \beta \ln(c_2) - \lambda * \left ( c_1 - \dfrac{c_2}{R} - c_2 - y_1 - \dfrac{y_2}{R} \right ) \implies \\ \dfrac{\delta L}{\delta c_1} = 0 \implies \dfrac{1}{c_1} - \lambda = 0; \\ \dfrac{\delta L}{\delta c_2} = 0 \implies \dfrac{\beta}{c_2} - \lambda = 0; \text { and }\\ \dfrac{\delta L}{\delta \lambda} = 0 \implies c_1 - \dfrac{c_2}{R} - y_1 - \dfrac{y_2}{R}.$$
Eliminating lambda from the first two partials gives

$$\dfrac{1}{c_1} - \dfrac{\beta}{c_2} = 0 \implies c_2 = \beta c_1.$$
But that does not match what you say the answer is.

In your very first equation should it be

$$u = c_1 + \dfrac{\beta c_2}{R}$$
please try to clarify the question.

 

[j4m]

It is hard to be sure that I am doing this correctly when I have no idea what any of the variables represent. In fact, you have not even bothered to tell us what are variables and what are constants. Moreover, the utility equation has a very strange symbol in it. How many variables are we dealing with? Is BC an abbreviation for Before Christ? We are not mind readers.

It looks like a utility maximization problem subject to a budget constraint.

Consider the problem: maximize u subject to the constraint that $c_1 - c_2/R - y_1 - y_2/R = 0$, where all variables are positive and y_1, y_2, and R are constants.

$$L(c_1, \ c_2) = \ln(c_1) + \beta \ln(c_2) - \lambda * \left ( c_1 - \dfrac{c_2}{R} - c_2 - y_1 - \dfrac{y_2}{R} \right ) \implies \\ \dfrac{\delta L}{\delta c_1} = 0 \implies \dfrac{1}{c_1} - \lambda = 0; \\ \dfrac{\delta L}{\delta c_2} = 0 \implies \dfrac{\beta}{c_2} - \lambda = 0; \text { and }\\ \dfrac{\delta L}{\delta \lambda} = 0 \implies c_1 - \dfrac{c_2}{R} - y_1 - \dfrac{y_2}{R}.$$
Eliminating lambda from the first two partials gives

$$\dfrac{1}{c_1} - \dfrac{\beta}{c_2} = 0 \implies c_2 = \beta c_1.$$
But that does not match what you say the answer is.

In your very first equation should it be

$$u = c_1 + \dfrac{\beta c_2}{R}$$
please try to clarify the question.

That "very strange symbol" is just my lecturer's cursor by a beta. I also referred to BC as the 'budget constraint' but perhaps I'll put it in big bold letters for you next time. Otherwise, thanks for the help buddy

 

[topsquark]

Otherwise, thanks for the help buddy

JeffM is a girl.

-Dan

 

[JeffM]

That "very strange symbol" is just my lecturer's cursor by a beta. I also referred to BC as the 'budget constraint' but perhaps I'll put it in big bold letters for you next time. Otherwise, thanks for the help buddy

Oh, you are quite welcome.