Answer:
l dont know.... but
Step-by-step explanation:
hopes it help
The Lagrange multipliers technique used to find the local maxima and minima of function.
The demand function of X and Y in general form is [tex]I=\dfrac{1}{a}\times x\times P_x[/tex].
Given:
Maximize [tex]U=a\log x+(1-a)\log y[/tex]
Find the derivative of given function with respect x.
[tex]mu_x= \dfrac {\partial u}{\partial x}=a\times \dfrac{1}{x}+(1-a)\times 0\\ mu_x=\dfrac{a}{x}[/tex]
Find the derivative of given function with respect Y.
[tex]mu_y= \dfrac {\partial u}{\partial y}=a\times 0+(1-a)\times \dfrac{1}{y}\\ mu_x=\dfrac{1-a}{y}[/tex]
Now equate the above two equation.
[tex]\dfrac {mu_x}{P_x}=\dfrac{mu_y}{P_y}[/tex]
Substitute the value.
[tex]\begin{aligned}\dfrac{\frac{a}{x}}{P_x}&=\dfrac{\frac{1-a}{y}}{P_y}\\\dfrac{a}{x\times P_x}&=\dfrac{(1-y)}{y\times P_y}\\a_y\times P_y&=(1-a)\times P_x\\y&=\dfrac{1-a}{a}\times x\times \dfrac{P_x}{P_y}\\\end[/tex]
As per the question,
[tex]I=P\times x+P\times y[/tex]
Substitute the value of y.
[tex]I=P\times x+P\times \dfrac{1-a}{a}\times x\times \dfrac{P_x}{P_y}\\I=x\times P_x\times \left (\dfrac{a+1-1}{a}\right)\\I=\dfrac{1}{a}\times x\times P_x[/tex]
Thus, the demand function of X and Y in general form is [tex]I=\dfrac{1}{a}\times x\times P_x[/tex].
Learn more about Lagrange multipliers here:
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