Thread: Maximizing Utility through Lagrange Multiplier

1. 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) ?

2. Hello,

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 ?
Do that

But I must say... this is quite nasty

3. Originally Posted by 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 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

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

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

Then we have

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

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

Finally we have

$\left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\8X + 16Y - 480 = 0; \hfill \\ \end{gathered} \right.$ $\Leftrightarrow \left\{ \begin{gathered}X = \frac{Y}{2}, \hfill \\20Y = 480; \hfill \\ \end{gathered} \right.$ $\Leftrightarrow \left\{ \begin{gathered}X = 12, \hfill \\Y = 24. \hfill \\ \end{gathered} \right.$

4. 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 ?

5. 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]

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

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]: . $y \:=\:2(12) \quad\Rightarrow\quad \boxed{y\:=\:24}$

6. 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 $5X = 60 \implies X = 12 ,~ Y = 24$

7. Originally Posted by Soroban
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 I’m a little bit lacking in the manipulation of algebra, and I can’t 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 can’t figure out how to rearrange, is there a simple way in which I’m just overlooking ?

8. For the first:

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

Second:

$
-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,
$-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 $\lambda$ and got -0.04 ?

10. Originally Posted by Apache
Thanks, i think ive cracked it, solved the equation for $\lambda$ and got -0.04 ?
Yes, $\lambda = - \frac{1}{{20}}\sqrt[5]{{\frac{1}{2}}} = - \frac{{\sqrt[5]{{16}}}}{{40}} \approx - 0.044$.