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Math Help - Maximizing Utility through Lagrange Multiplier

  1. #1
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    Maximizing Utility through Lagrange Multiplier

    A consumers utility function is given by  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 U=X^{0.2}Y^{0.8} subject to constraint 8X + 16Y = 480

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

    Partially differentiating gives

    \frac{\partial V}{\partial X} = 0.2Y^{0.8}X^{-0.8}+\lambda8 (1)
    \frac{\partial V}{\partial Y} = 0.8Y^{-0.2}X^{0.2}+\lambda16 (2)
    \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
    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; July 28th 2009 at 08:03 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
    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
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Apache View Post
    A consumers utility function is given by  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 U=X^{0.2}Y^{0.8} subject to constraint 8X + 16Y = 480

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

    Partially differentiating gives

    \frac{\partial V}{\partial X} = 0.2Y^{0.8}X^{-0.8}+\lambda8 (1)
    \frac{\partial V}{\partial Y} = 0.8Y^{-0.2}X^{0.2}+\lambda16 (2)
    \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
    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

    \left\{ \begin{gathered}<br />
  0.2{Y^{0.8}}{X^{ - 0.8}} + 8\lambda  = 0, \hfill \\<br />
  0.8{Y^{ - 0.2}}{X^{0.2}} + 16\lambda  = 0; \hfill \\ <br />
\end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}\frac{1}{5}\frac{{{Y^{\frac{4}{5}}  }}}{{{X^{\frac{4}<br />
{5}}}}} + 8\lambda  = 0, \hfill \\\frac{4}<br />
{5}\frac{{{X^{\frac{1}{5}}}}}{{{Y^{\frac{1}<br />
{5}}}}} + 16\lambda  = 0; \hfill \\ \end{gathered}  \right. \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.

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

    Then we have

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

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

    Finally we have

    \left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\8X + 16Y - 480 = 0; \hfill \\ \end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\20Y = 480; \hfill \\ \end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}X = 12, \hfill \\Y = 24. \hfill \\ \end{gathered}  \right.
    Last edited by DeMath; July 28th 2009 at 10:29 AM.
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  4. #4
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    Thank you,

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

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

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

    Rearranging (1) and (2) gives,

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

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

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

    0.4Y^{0.8}X^{-0.8} = 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; July 28th 2009 at 10:12 AM.
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  5. #5
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    Hello, Apache!

    Hello, Apache!

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

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


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

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

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


    Partially differentiating gives

    . . \begin{array}{cccccc}\dfrac{\partial V}{\partial X} &=&  0.2Y^{0.8}X^{-0.8}+8\lambda &=& 0 & {\color{blue}[1]} \\ \\[-3mm]<br /> <br />
\dfrac{\partial V}{\partial Y} &=&  0.8Y^{-0.2}X^{0.2}+16\lambda &=& 0 & {\color{blue}[2]} \\ \\[-3mm]<br /> <br />
\dfrac{\partial U}{\partial \lambda} &=&  8X + 16Y - 480 &=& 0 & {\color{blue}(3)} \end{array}
    I solved it like this . . .


    Multiply [1] by 5: . 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: . 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]: . -\frac{1}{40}\left(\frac{y}{x}\right)^{0.8} \;=\;-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2}

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

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

    Substitute into [5]: . 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

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

    let  P = X^{0.2} Y^{0.8}

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

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

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

     = 16 ( 12 )^5

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

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

    so  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: . 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: . 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]: . -\frac{1}{40}\left(\frac{y}{x}\right)^{0.8} \;=\;-\frac{1}{20}\left(\frac{x}{y}\right)^{0.2}

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

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

    Substitute into [5]: . 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
    x^{-0.8}y^{0.8} \!+ 40\lambda \:=\:0
    becomes
     \quad\Rightarrow\quad \left(\frac{y}{x}\right)^{0.8} + 40\lambda \:=\:0
    And similarly when
    -40\left(\frac{y}{x}\right)^{0.2}
    I fail to see how that gets
    :\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:

    <br />
x^{-0.8}y^{0.8} = \frac{y^{0.8}}{x^{0.8}}= \left(\frac{y}{x}\right)^{0.8}<br />

    Second:

    <br />
-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}<br />

    Also,
    -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<br />
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  9. #9
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    Thanks, i think ive cracked it, solved the equation for \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 \lambda and got -0.04 ?
    Yes, \lambda  =  - \frac{1}{{20}}\sqrt[5]{{\frac{1}{2}}} =  - \frac{{\sqrt[5]{{16}}}}{{40}} \approx  - 0.044.
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