Stone-Geary utility maximization for a given budget

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Paul Verheyen

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Created this as a copy of Nut_maxU_2020.

Name	Status	Author	Last Modified
Demand_functions_from_ a_ utility_function	Ready to use	Paul Verheyen	15/04/2020 13:08
Stone-Geary utility and calculation of elasticities	Ready to use	Paul Verheyen	15/04/2020 11:36
Stone-Geary utility maximization for a given budget	Ready to use	Paul Verheyen	10/10/2024 08:33
Budget minimization for a given Stone-Geary utility function	Ready to use	Paul Verheyen	10/10/2024 08:35

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Question statement

An individual has as utility function

$U(q_1,q_2)={\var{c1}} \cdot (q_1-{\var{a1}})^{{1}/{\var{n1}}} \cdot (q_2-{\var{a2}})^{{\var{n1min1}}/{\var{n1}}}$

where $q_1$ and $q_2$ denote the quantities consumed of respectively the products 1 and 2.
The unit prices for this products are respectively $p_1 = \var{p1}$ and $p_2 = \var{p2}$ .
Maximize the utility for this individual when the available budget is {ybudget} EURO
(suppose the budget is completetely spent on these products).

Give any introductory information the student needs.

			Name	Type	Generated Value

n1	integer	9
theta	integer	3
c1	integer	3
s	integer	2
gegp2	integer	15
t1	integer	1
v	integer	7
eta	integer	30
n1min1	integer	8
a1	integer	27
a2	integer	540
t2	integer	8
n2	integer	9
n1min1maala1	integer	216
a1maalt2	integer	216
gegp1	integer	30
gegy	integer	10800
q1	integer	34
q2	integer	652
ybudget	integer	10800
p1	integer	30
p2	integer	15
y	integer	10800
nutfinaal	number	246.9153947023
u	number	246.9153947023
umin	number	82.3051315674
umax	number	411.5256578372
cu	number	142.8417879681
cumax	number	253.7669856566
cumin	number	41.5043738299
ydeelp2	rational	720
ydeelp1	rational	360

Automatically regenerate variables?

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Value

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

→ Used by:

a1
a2
cu
cumax
cumin
eta
gegy
n1min1
n2
nutfinaal
q2
t2

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Advice

Step 1: Formulate the corresponding maximum problem using the budget restriction.
\begin{eqnarray}
\max_{q_1,q_2} U(q_1,q_2) &=& {\var{c1}} \cdot (q_1-{\var{a1}})^{1 / {\var{n1}}} \cdot (q_2-{\var{a2}})^{{\var{n1min1}} / {\var{n1}}}
\mbox{ if } && \var{p1} \cdot q_1 + \var{p2} \cdot q_2 = \var{y}
\end{eqnarray}
or equivalently
\begin{eqnarray}
\max_{q_1,q_2} ln U(q_1,q_2) &=& ln({\var{c1}}) + {\frac{1}{\var{n1}}} \cdot ln(q_1-{\var{a1}}) + {\frac{\var{n1min1}}{\var{n1}}} \cdot ln(q_2-{\var{a2}}) \\
\mbox{ if } && \var{p1} \cdot q_1 + \var{p2} \cdot q_2 = \var{y}
\end{eqnarray}

Step 2: Write down the corresponding Lagrange function for this constrained maximum problem.
\begin{eqnarray}
L(q_1, q_2, \lambda) &=& ln U(q_1,q_2) + \lambda \cdot (y - p_1 \cdot q_1 - p_2 \cdot q_2) \\
&=& ln({\var{c1}}) + {\frac{1}{\var{n1}}} \cdot ln(q_1-{\var{a1}}) + {\frac{\var{n1min1}}{\var{n1}}} \cdot ln(q_2-{\var{a2}}) + \lambda \cdot ( \var{y} - \var{p1} \cdot q_1 - \var{p2}\cdot q_2 )
\end{eqnarray}

Step 3: Determine the first order conditions in order to calculate critical points of the Lagrangian.
\begin{eqnarray}
\mbox{(1) } \quad \frac{\partial {L}}{\partial q_1} & = & {\frac{1}{\var{n1}}} \cdot {\frac {1} {q_1-{\var{a1}}} } - \lambda \cdot \var{p1} = 0 \\
\mbox{(2) } \quad \frac{\partial {L}}{\partial q_2} & = & {\frac{\var{n1min1}}{\var{n1}}} \cdot {\frac {1} {q_2-{\var{a2}}}} - \lambda \cdot \var{p2} = 0 \\
\mbox{(3) } \quad \frac{\partial {L}}{\partial \lambda} & = & \var{y} - \var{p1} \cdot q_1 - \var{p2} \cdot q_2 = 0
\end{eqnarray}

Step 4: Find the solution of this system of equations in $q_1$, $q_2$ and $\lambda$.
Calculating $\lambda$ in both equations (1) and (2) leads to
\begin{eqnarray}
{\frac{1}{\var{n1}}}\cdot{\frac {1} {\var{p1}} \cdot (q_1-{\var{a1}}) }& = & {\frac{\var{n1min1}}{\var{n1}}}\cdot{\frac {1} {\var{p2} \cdot(q_2-{\var{a2}})}}\\
{\var{p2}} \cdot {(q_2-{\var{a2}})} & = & {\var{n1min1}} \cdot {\var{p1}} \cdot {(q_1-{\var{a1}})} \\
{\var{p2}} \cdot {q_2} & = & {\var{a2}} \cdot {\var{p2}} + {\var{n1min1}} \cdot {\var{p1}} \cdot {(q_1-{\var{a1}})} = {\var{a2*p2}} + {\var{n1min1*p1}} \cdot {(q_1-{\var{a1}})}
\end{eqnarray}
i.e.
\begin{eqnarray}
\mbox{(4)} \quad \quad \quad \quad \quad \quad \quad \quad{q_2} & = & {\var{a2-n1min1*s*a1}} + {\var{n1min1*s}} \cdot {q_1}
\end{eqnarray}
Substitute(4) in the budget restriction (3):
\begin{eqnarray}
{\var{p1}} \cdot {q_1} + {\var{(a2-n1min1*s*a1)*p2}} + {\var{n1min1*s*p2}} \cdot {q_1} & = & \var{y}\\
{q_1} + {\var{a2-n1min1*a1}} + {\var{n1min1}} \cdot {q_1} & = & \var{n1*v+a1+a2}\\
{\var{n1}} \cdot {q_1} + {\var{a2-n1min1*a1}} & = & \var{n1*v+a1+a2}\\
q^*_1 &=& \var{q1}
\end{eqnarray}
Substituting this last formula in equation (4), we can calculate $q_2$:
\begin{eqnarray}
{q^*_2} & = & \var{q2}
\end{eqnarray}

Answer.

For
\begin{eqnarray}
q^*_1&=&\var{q1} \\
q^*_2&=&\var{q2}
\end{eqnarray}
the maximal utility value is
\[ U (q^*_1,q^*_2) \approx \var{nutfinaal} \]

{toonfiguur()}

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Stone-Geary utility maximization for a given budget

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Paul Verheyen 4 years, 11 months ago

Paul Verheyen 4 years, 11 months ago

Paul Verheyen 4 years, 11 months ago

An individual has as utility function

$U(q_1,q_2)={\var{c1}} \cdot (q_1-{\var{a1}})^{{1}/{\var{n1}}} \cdot (q_2-{\var{a2}})^{{\var{n1min1}}/{\var{n1}}}$

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Generated value: `integer`

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Determine the optimal value of $q_1$:

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Stone-Geary utility maximization for a given budget

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Comment

Checkpoint description

Paul Verheyen 4 years, 11 months ago

Paul Verheyen 4 years, 11 months ago

Paul Verheyen 4 years, 11 months ago

An individual has as utility function

U(q1,q2)={c1}⋅(q1−{a1})1/{n1}⋅(q2−{a2}){n1min1}/{n1} U(q_1,q_2)={\var{c1}} \cdot (q_1-{\var{a1}})^{{1}/{\var{n1}}} \cdot (q_2-{\var{a2}})^{{\var{n1min1}}/{\var{n1}}}

Error in variable testing condition

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Generated value: integer

Parts

Number entry

Determine the optimal value of $q_1$:

Advanced settings

$U(q_1,q_2)={\var{c1}} \cdot (q_1-{\var{a1}})^{{1}/{\var{n1}}} \cdot (q_2-{\var{a2}})^{{\var{n1min1}}/{\var{n1}}}$

Generated value: `integer`