nacho-man
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Assuming $x_1, x_2 \geq 0, \lambda \neq 0, w_1,w_2 > 0$
We have the equalities:
$$w_1 - \lambda x_2 = 0 ... (1)$$
$$w_2 - \lambda x_1 = 0 ... (2)$$
$$\bar y - x_1x_2= 0 ... (3)$$
My solutions say that $\lambda^* = \sqrt\frac{w_1w_2}{\bar y}$
Which I was able to solve myself.
The other solution is $x^* = (\frac{w_2 \bar y}{w_1}, \frac{w_1 \bar y}{w_2})$
Which I cannot seem to get. Would anyone be so kind as to point out how to obtain solutions for $x^*$ ?
I tried:
$x_1= \frac{w_2}{\lambda}$ and $x_2 = \frac{w_1}{\lambda}$
then put these into the 3rd equation, but ended up gettig
$\bar y = \frac{w_1w_2}{\lambda^2}$ which I couldn't see turning into what I needed either way - even just subbing in one of the x-values at a time, but to no avail.
We have the equalities:
$$w_1 - \lambda x_2 = 0 ... (1)$$
$$w_2 - \lambda x_1 = 0 ... (2)$$
$$\bar y - x_1x_2= 0 ... (3)$$
My solutions say that $\lambda^* = \sqrt\frac{w_1w_2}{\bar y}$
Which I was able to solve myself.
The other solution is $x^* = (\frac{w_2 \bar y}{w_1}, \frac{w_1 \bar y}{w_2})$
Which I cannot seem to get. Would anyone be so kind as to point out how to obtain solutions for $x^*$ ?
I tried:
$x_1= \frac{w_2}{\lambda}$ and $x_2 = \frac{w_1}{\lambda}$
then put these into the 3rd equation, but ended up gettig
$\bar y = \frac{w_1w_2}{\lambda^2}$ which I couldn't see turning into what I needed either way - even just subbing in one of the x-values at a time, but to no avail.