# Thread: Maximizing Utility through Lagrange Multiplier

1. ## 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) ?

2. Hello,

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 ?
Do that

But I must say... this is quite nasty

3. Originally Posted by 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 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.$

4. 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 ?

5. 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}$6. A non-calculus solution , we have$\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$7. Originally Posted by Soroban Hello, Apache! Hello, Apache! I solved it like this . . . Multiply [1] by 5: .$\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}$I understand the method used however, I find I’m a little bit lacking in the manipulation of algebra, and I can’t get my head around how$\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$I just can’t figure out how to rearrange, is there a simple way in which I’m just overlooking ? 8. For the first:$\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
$9. Thanks, i think ive cracked it, solved the equation for$\displaystyle \lambda$and got -0.04 ? 10. Originally Posted by Apache Thanks, i think ive cracked it, solved the equation for$\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\$.