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# System of equations

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Normally I would give some context, but I highly doubt whether that is necessary. (Since it might only complicate the problem). I have this system of equations

$$\begin{array}{lcl} \frac{1}{2}x_1^{-\frac{1}{2}}x_2^{\frac{1}{3}}- \lambda p_1 = 0\\ \frac{1}{3}x_1^{\frac{1}{2}}x_2^{-\frac{2}{3}}- \lambda p_2 = 0\\ p_1x_1+p_2x_2 = m \end{array}$$

of which I want the functions x(p,m) for x1 and x2.

It is given in the answers that

$$\begin{array}{lcl} x_1 = \frac{3}{5} \frac{m}{p_1}\\ x_2 = \frac{2}{5} \frac{m}{p_2}\\ \end{array}$$

but I want to know how they got there.

I know the answer is always of the form

$$\begin{array}{lcl} x_1 = a_1 \frac{m}{p_1}\\ x_2 = a_2 \frac{m}{p_2}\\ \end{array}$$

where

$$a_1+a_2 = 1$$

Since

$$p_1a_1 \frac{m}{p_1} + p_2a_2 \frac{m}{p_2} = m \Rightarrow (a_1+a_2)m = m \Rightarrow a_1+a_2 = 1$$

So perhaps this can be used to find values for a1 and a2, but I keep getting stuck.

Reinout

Jun 22, 2014

#3
+11

$$\\ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } =\lambda p_1\\\\ \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\lambda p_2$$

$$\dfrac{ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } { \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\dfrac{\lambda p_1}{\lambda p_2}$$

$$\Rightarrow (1) \quad \dfrac {p_1}{p_2}=\dfrac{3}{2}\dfrac{x_2}{x_1}$$

$$\begin{array}{rcccc} &I.&&II.&\\\\ (2)\qquad p_1x_1+p_2x_2=m \quad | \quad &\times&\dfrac{1}{p_2} \quad | \quad &\times&\dfrac{1}{p_1} \end{array}$$

$$I.\quad \dfrac{p_1}{p_2}x_1+x_2 = \dfrac{m}{p_2} \quad \Rightarrow \quad \dfrac{3}{2}\dfrac{x_2}{x_1}x_1+x_2=\dfrac{m}{p_2}\quad \Rightarrow \quad \dfrac{5}{2}x_2=\dfrac{m}{p_2}$$

$${\boxed{x_2=\dfrac{2}{5}\dfrac{m}{p_2}}}$$

$$II.\quad x_1+\dfrac{p_2}{p_1}x_2 = \dfrac{m}{p_1}\quad\Rightarrow \quad x_1+ \dfrac{2}{3}\dfrac{x_1}{x_2}x_2=\dfrac{m}{p_1} \quad \Rightarrow \quad \dfrac{5}{3}x_1=\dfrac{m}{p_1}$$

$${\boxed{x_1=\dfrac{3}{5}\dfrac{m}{p_1}} }$$

.
Jun 23, 2014

#1
+5

Multiply the first equation by x1^(1/2), and the second by x2^(2/3).

Move the negative terms to the other side in each equation, and then align them so that both x1 terms are on the same side, (the lhs say, the x2 terms on the rhs).

The product of the two lhs's will equal the product of the two rhs's, do this and simplify.

That should get you 3x2p2 = 2x1p1

Now solve in conjunction with the third equation.

Jun 22, 2014
#2
+10

Here's my take on this:  Jun 23, 2014
#3
+11

$$\\ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } =\lambda p_1\\\\ \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\lambda p_2$$

$$\dfrac{ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } { \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\dfrac{\lambda p_1}{\lambda p_2}$$

$$\Rightarrow (1) \quad \dfrac {p_1}{p_2}=\dfrac{3}{2}\dfrac{x_2}{x_1}$$

$$\begin{array}{rcccc} &I.&&II.&\\\\ (2)\qquad p_1x_1+p_2x_2=m \quad | \quad &\times&\dfrac{1}{p_2} \quad | \quad &\times&\dfrac{1}{p_1} \end{array}$$

$$I.\quad \dfrac{p_1}{p_2}x_1+x_2 = \dfrac{m}{p_2} \quad \Rightarrow \quad \dfrac{3}{2}\dfrac{x_2}{x_1}x_1+x_2=\dfrac{m}{p_2}\quad \Rightarrow \quad \dfrac{5}{2}x_2=\dfrac{m}{p_2}$$

$${\boxed{x_2=\dfrac{2}{5}\dfrac{m}{p_2}}}$$

$$II.\quad x_1+\dfrac{p_2}{p_1}x_2 = \dfrac{m}{p_1}\quad\Rightarrow \quad x_1+ \dfrac{2}{3}\dfrac{x_1}{x_2}x_2=\dfrac{m}{p_1} \quad \Rightarrow \quad \dfrac{5}{3}x_1=\dfrac{m}{p_1}$$

$${\boxed{x_1=\dfrac{3}{5}\dfrac{m}{p_1}} }$$

heureka Jun 23, 2014
#4
0

Thank you guys,

Since I'm doing a resit it took me a while to get back to this question, but your answers are really helpful I hope I  will pass this time Jul 4, 2014