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Thread: Maximizing Utility through Lagrange Multiplier

  1. #1
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    Maximizing Utility through Lagrange Multiplier

    A consumers utility function is given by $\displaystyle U = X^{0.2}Y^{0.8}$
    The consumer has a budget of £480, the price of good X and good Y are £8 and £16 per unit.

    Using the Lagrange multiplier method, identify the quantities of X and Y the consumer will purchase in order to maximise their utility?
    This is what I have so far...

    Px =8 Py=16 B=480

    Maximise $\displaystyle U=X^{0.2}Y^{0.8}$ subject to constraint 8X + 16Y = 480

    $\displaystyle V=X^{0.2}Y^{0.8} + \lambda[8X + 16Y -480]$

    Partially differentiating gives

    $\displaystyle \frac{\partial V}{\partial X}$ = $\displaystyle 0.2Y^{0.8}X^{-0.8}+\lambda8$ (1)
    $\displaystyle \frac{\partial V}{\partial Y}$ = $\displaystyle 0.8Y^{-0.2}X^{0.2}+\lambda16$ (2)
    $\displaystyle \frac{\partial V}{\partial \lambda} $ = 8X + 16Y – 480 (3)

    I’ve then set these 3 equations equal to zero, in order to solve them.

    However I’m uncertain as to whether I have to multiply equation (1) which would give me
    $\displaystyle 0.4Y^{0.8}X^{-0.8}+\lambda16$ and then set this equal to equation 2 ?

    Or do I just divide equation (1) by equation (2) ?
    Last edited by Apache; Jul 28th 2009 at 07:03 AM.
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  2. #2
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    Hello,

    However Iím uncertain as to whether I have to multiply equation (1) which would give me
    $\displaystyle 0.4Y^{0.8}X^{-0.8}+\lambda16$ and then set this equal to equation 2 ?
    Do that


    But I must say... this is quite nasty
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  3. #3
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    Quote Originally Posted by Apache View Post
    A consumers utility function is given by $\displaystyle U = X^{0.2}Y^{0.8}$
    The consumer has a budget of £480, the price of good X and good Y are £8 and £16 per unit.

    Using the Lagrange multiplier method, identify the quantities of X and Y the consumer will purchase in order to maximise their utility?
    This is what I have so far...

    Px =8 Py=16 B=480

    Maximise $\displaystyle U=X^{0.2}Y^{0.8}$ subject to constraint 8X + 16Y = 480

    $\displaystyle V=X^{0.2}Y^{0.8} + \lambda[8X + 16Y -480]$

    Partially differentiating gives

    $\displaystyle \frac{\partial V}{\partial X}$ = $\displaystyle 0.2Y^{0.8}X^{-0.8}+\lambda8$ (1)
    $\displaystyle \frac{\partial V}{\partial Y}$ = $\displaystyle 0.8Y^{-0.2}X^{0.2}+\lambda16$ (2)
    $\displaystyle \frac{\partial V}{\partial \lambda} $ = 8X + 16Y – 480 (3)

    I’ve then set these 3 equations equal to zero, in order to solve them.

    However I’m uncertain as to whether I have to multiply equation (1) which would give me
    $\displaystyle 0.4Y^{0.8}X^{-0.8}+\lambda16$ and then set this equal to equation 2 ?

    Or do I just divide equation (1) by equation (2) ?
    First, consider the first two equations

    $\displaystyle \left\{ \begin{gathered}
    0.2{Y^{0.8}}{X^{ - 0.8}} + 8\lambda = 0, \hfill \\
    0.8{Y^{ - 0.2}}{X^{0.2}} + 16\lambda = 0; \hfill \\
    \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}\frac{1}{5}\frac{{{Y^{\frac{4}{5}} }}}{{{X^{\frac{4}
    {5}}}}} + 8\lambda = 0, \hfill \\\frac{4}
    {5}\frac{{{X^{\frac{1}{5}}}}}{{{Y^{\frac{1}
    {5}}}}} + 16\lambda = 0; \hfill \\ \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}{\left( {{{\left( {\frac{X}{Y}} \right)}^{^{\frac{1}{5}}}}} \right)^{ - 4}} + 40\lambda = 0, \hfill \\{\left( {\frac{X}{Y}} \right)^{^{\frac{1}{5}}}} + 20\lambda = 0. \hfill \\ \end{gathered} \right.$

    $\displaystyle {\text{Let }}{\left( {\frac{X}{Y}} \right)^{{1 \mathord{\left/
    {\vphantom {1 5}} \right.\kern-\nulldelimiterspace} 5}}} = t \Rightarrow {\left( {{{\left( {\frac{X}{Y}} \right)}^{{1 \mathord{\left/
    {\vphantom {1 5}} \right.\kern-\nulldelimiterspace} 5}}}} \right)^{ - 4}} = \frac{1}{{{t^4}}}.$

    Then we have

    $\displaystyle \left\{ \begin{gathered}\frac{1}{{{t^4}}} + 40\lambda = 0, \hfill \\
    t + 20\lambda = 0; \hfill \\ \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}
    \frac{1}
    {{{t^4}}} - 2t = 0, \hfill \\
    \lambda = - \frac{t}
    {{20}}; \hfill \\
    \end{gathered} \right. $$\displaystyle \Leftrightarrow \left\{ \begin{gathered}{t^5} = \frac{1}
    {2}, \hfill \\
    \lambda = - \frac{t}
    {{20}}; \hfill \\
    \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}
    t = \sqrt[5]{{\frac{1}
    {2}}}, \hfill \\
    \lambda = - \frac{1}
    {{20}}\sqrt[5]{{\frac{1}
    {2}}}. \hfill \\
    \end{gathered} \right.$

    $\displaystyle {\text{So }}{\left ( {\frac{X}{Y}} \right)^{{1 \mathord{\left/
    {\vphantom {1 5}} \right.\kern-\nulldelimiterspace} 5}}} = \sqrt[5]{{\frac{1}{2}}} \Leftrightarrow \frac{X}{Y} = \frac{1}
    {2} \Leftrightarrow X = \frac{Y}{2}.$

    Finally we have

    $\displaystyle \left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\8X + 16Y - 480 = 0; \hfill \\ \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\20Y = 480; \hfill \\ \end{gathered} \right.$$\displaystyle \Leftrightarrow \left\{ \begin{gathered}X = 12, \hfill \\Y = 24. \hfill \\ \end{gathered} \right.$
    Last edited by DeMath; Jul 28th 2009 at 09:29 AM.
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  4. #4
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    Thank you,

    After multiplying equation (1) by 2 I get that to be

    $\displaystyle 0.4Y^{0.8}X^{-0.8}+\lambda16$

    Equation (2) is $\displaystyle 0.8Y^{-0.2}X^{0.2}+\lambda16$

    Rearranging (1) and (2) gives,

    (1) $\displaystyle 0.4Y^{0.8}X^{-0.8}= - \lambda16$

    (2) $\displaystyle 0.8Y^{-0.2}X^{0.2}= - \lambda16$

    I then set equation (1) equal to equation (2) which gives,

    $\displaystyle 0.4Y^{0.8}X^{-0.8}$ = $\displaystyle 0.8Y^{-0.2}X^{0.2}$

    I guess I then have to try and rearrange that somehow to come up with a solution of Y ?
    Last edited by Apache; Jul 28th 2009 at 09:12 AM.
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  5. #5
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    Hello, Apache!

    Hello, Apache!

    A consumers utility function is given by: $\displaystyle U \:=\:x^{0.2}y^{0.8}$
    The consumer has a budget of $480, the price of goods $\displaystyle x$ and $\displaystyle y$ are $8 and $16 per unit.

    Using the Lagrange multiplier method, identify the quantities $\displaystyle x$ and $\displaystyle y$
    the consumer will purchase in order to maximise their utility.


    This is what I have so far: .$\displaystyle P_x =8 \quad P_y = 6\quad B =480$

    Maximize $\displaystyle U\:=\:x^{0.2}y^{0.8}$ subject to constraint $\displaystyle 8x + 16y \:=\: 480$

    . . $\displaystyle U \;=\; x^{0.2}y^{0.8} + \lambda(8X + 16Y -480)$


    Partially differentiating gives

    . . $\displaystyle \begin{array}{cccccc}\dfrac{\partial V}{\partial X} &=& 0.2Y^{0.8}X^{-0.8}+8\lambda &=& 0 & {\color{blue}[1]} \\ \\[-3mm]

    \dfrac{\partial V}{\partial Y} &=& 0.8Y^{-0.2}X^{0.2}+16\lambda &=& 0 & {\color{blue}[2]} \\ \\[-3mm]

    \dfrac{\partial U}{\partial \lambda} &=& 8X + 16Y - 480 &=& 0 & {\color{blue}(3)} \end{array}$
    I solved it like this . . .


    Multiply [1] by 5: .$\displaystyle x^{-0.8}y^{0.8} \!+ 40\lambda \:=\:0 \quad\Rightarrow\quad \left(\frac{y}{x}\right)^{0.8} + 40\lambda \:=\:0 \quad\Rightarrow\quad \lambda \:=\:-\frac{1}{40}\left(\frac{y}{x}\right)^{0.8}\;\;{\co lor{blue}[3]}$

    Multiply [2] by 5: .$\displaystyle 4x^{0.2}y^{-0.2} + 80\lambda \:=\:0 \quad\Rightarrow\quad 4\left(\frac{x}{y}\right)^{0.2}\!\! + 80\lambda \:=\:0 \quad\Rightarrow\quad \lambda \:=\:-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2} \;\;{\color{blue}[4]}$

    Equate [3] and [4]: .$\displaystyle -\frac{1}{40}\left(\frac{y}{x}\right)^{0.8} \;=\;-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2}$

    Multiply by $\displaystyle -40\left(\frac{y}{x}\right)^{0.2}\!:\quad\frac{y}{x } \:=\:2 \quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[5]}$

    Substitute into [3]: . $\displaystyle 8x + 16(2x) - 480 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:12}$

    Substitute into [5]: . $\displaystyle y \:=\:2(12) \quad\Rightarrow\quad \boxed{y\:=\:24}$

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  6. #6
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    A non-calculus solution ,

    we have

    $\displaystyle 8X + 16Y = 480 \implies X + 2Y = 60 $

    let $\displaystyle P = X^{0.2} Y^{0.8} $

    $\displaystyle P^5 = X Y^4 = 2^4 X (\frac{Y}{2})^4$

    $\displaystyle = 2^4 X ( \frac{Y}{2} ) ( \frac{Y}{2} ) ( \frac{Y}{2} ) ( \frac{Y}{2} ) $

    $\displaystyle \leq 2^4 [ \frac{ X + 4 \frac{Y}{2}}{5} ]^5 = 16 [ \frac{ X + 2Y}{5} ]^5 $

    $\displaystyle = 16 ( 12 )^5 $

    so $\displaystyle P \leq 12(2^{4/5})$

    the equality holds when $\displaystyle X = \frac{Y}{2} $

    so $\displaystyle 5X = 60 \implies X = 12 ,~ Y = 24$
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, Apache!

    Hello, Apache!

    I solved it like this . . .


    Multiply [1] by 5: .$\displaystyle x^{-0.8}y^{0.8} \!+ 40\lambda \:=\:0 \quad\Rightarrow\quad \left(\frac{y}{x}\right)^{0.8} + 40\lambda \:=\:0 \quad\Rightarrow\quad \lambda \:=\:-\frac{1}{40}\left(\frac{y}{x}\right)^{0.8}\;\;{\co lor{blue}[3]}$

    Multiply [2] by 5: .$\displaystyle 4x^{0.2}y^{-0.2} + 80\lambda \:=\:0 \quad\Rightarrow\quad 4\left(\frac{x}{y}\right)^{0.2}\!\! + 80\lambda \:=\:0 \quad\Rightarrow\quad \lambda \:=\:-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2} \;\;{\color{blue}[4]}$

    Equate [3] and [4]: .$\displaystyle -\frac{1}{40}\left(\frac{y}{x}\right)^{0.8} \;=\;-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2}$

    Multiply by $\displaystyle -40\left(\frac{y}{x}\right)^{0.2}\!:\quad\frac{y}{x } \:=\:2 \quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[5]}$

    Substitute into [3]: . $\displaystyle 8x + 16(2x) - 480 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:12}$

    Substitute into [5]: . $\displaystyle y \:=\:2(12) \quad\Rightarrow\quad \boxed{y\:=\:24}$
    I understand the method used however, I find Iím a little bit lacking in the manipulation of algebra, and I canít get my head around how
    $\displaystyle x^{-0.8}y^{0.8} \!+ 40\lambda \:=\:0$
    becomes
    $\displaystyle \quad\Rightarrow\quad \left(\frac{y}{x}\right)^{0.8} + 40\lambda \:=\:0 $
    And similarly when
    $\displaystyle -40\left(\frac{y}{x}\right)^{0.2}$
    I fail to see how that gets
    $\displaystyle :\quad\frac{y}{x} \:=\:2$

    I just canít figure out how to rearrange, is there a simple way in which Iím just overlooking ?
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  8. #8
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    For the first:

    $\displaystyle
    x^{-0.8}y^{0.8} = \frac{y^{0.8}}{x^{0.8}}= \left(\frac{y}{x}\right)^{0.8}
    $

    Second:

    $\displaystyle
    -40 \cdot \left(-\frac{1}{40}\right)\left(\frac{y}{x}\right)^{0.8}\ left(\frac{y}{x}\right)^{0.2} = \left(\frac{y}{x}\right)^{0.8+0.2}=\frac{y}{x}
    $

    Also,
    $\displaystyle -40\cdot\left(-\frac{1}{20}\right)\left(\frac{x}{y}\right)^{0.2}\ left(\frac{y}{x}\right)^{0.2}=2\cdot\left(\frac{x} {y}\right)^{0.2}\left(\frac{x}{y}\right)^{-0.2}=2
    $
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  9. #9
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    Thanks, i think ive cracked it, solved the equation for $\displaystyle \lambda$ and got -0.04 ?
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  10. #10
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Apache View Post
    Thanks, i think ive cracked it, solved the equation for $\displaystyle \lambda$ and got -0.04 ?
    Yes, $\displaystyle \lambda = - \frac{1}{{20}}\sqrt[5]{{\frac{1}{2}}} = - \frac{{\sqrt[5]{{16}}}}{{40}} \approx - 0.044$.
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