ngihlungeqtr

2022-05-03

This is a utility maximzation problem

maximize ${x}^{a}+{y}^{b}$ subject to ${p}_{1}x+{p}_{2}y=w$ (utility maximization problem)

Anyone has any idea, there are no restrictions on $a$ and $b$, as far as i can see it. many thanks!!!

maximize ${x}^{a}+{y}^{b}$ subject to ${p}_{1}x+{p}_{2}y=w$ (utility maximization problem)

Anyone has any idea, there are no restrictions on $a$ and $b$, as far as i can see it. many thanks!!!

Alice Harmon

Beginner2022-05-04Added 12 answers

When consumers maximize utility $u(x,y)={x}^{a}+{y}^{b}$ with respect to a budget constraint ${p}_{1}x+{p}_{2}y=w$, the indifference curve is tangent to the budget line, therefore:

$MR{S}_{xy}={p}_{1}/{p}_{2}$

with $MR{S}_{xy}=\frac{{u}_{x}}{{x}_{y}}$ and

$\begin{array}{rl}{u}_{x}& =\frac{\mathrm{\partial}u(x,y)}{\mathrm{\partial}x}=a{x}^{a-1}\\ {u}_{y}& =\frac{\mathrm{\partial}u(x,y)}{\mathrm{\partial}y}=b{y}^{b-1}\end{array}$

that is

$\frac{{p}_{1}}{{p}_{2}}=\frac{a{x}^{a-1}}{b{y}^{b-1}}$

So the optimal $\hat{x}$ and $\hat{y}$ are solution of

$\begin{array}{rl}w& ={p}_{1}\hat{x}+{p}_{2}{\left(\frac{{p}_{2}}{{p}_{1}}\frac{a}{b}{\hat{x}}^{a-1}\right)}^{\frac{1}{b-1}}\\ w& ={p}_{1}{\left(\frac{{p}_{1}}{{p}_{2}}\frac{b}{a}{\hat{y}}^{b-1}\right)}^{\frac{1}{a-1}}+{p}_{2}\hat{y}\end{array}$

$MR{S}_{xy}={p}_{1}/{p}_{2}$

with $MR{S}_{xy}=\frac{{u}_{x}}{{x}_{y}}$ and

$\begin{array}{rl}{u}_{x}& =\frac{\mathrm{\partial}u(x,y)}{\mathrm{\partial}x}=a{x}^{a-1}\\ {u}_{y}& =\frac{\mathrm{\partial}u(x,y)}{\mathrm{\partial}y}=b{y}^{b-1}\end{array}$

that is

$\frac{{p}_{1}}{{p}_{2}}=\frac{a{x}^{a-1}}{b{y}^{b-1}}$

So the optimal $\hat{x}$ and $\hat{y}$ are solution of

$\begin{array}{rl}w& ={p}_{1}\hat{x}+{p}_{2}{\left(\frac{{p}_{2}}{{p}_{1}}\frac{a}{b}{\hat{x}}^{a-1}\right)}^{\frac{1}{b-1}}\\ w& ={p}_{1}{\left(\frac{{p}_{1}}{{p}_{2}}\frac{b}{a}{\hat{y}}^{b-1}\right)}^{\frac{1}{a-1}}+{p}_{2}\hat{y}\end{array}$

Olive Guzman

Beginner2022-05-05Added 16 answers

You could use Lagrange multipliers but I think that the problem could be easy if you extract "y" from the constraint and plug it in the objective function which becomes only dependent on "x". Now, serach for the maximum.

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