E Laurie

2022-10-02

USE THIS REFERENCE FOR EACH LETTER:

a.) L =

b.) FOC (x1) =

c.) FOC (x2) =

d.) FOC(lambda) =

e.) x1 =

f.) x2

g.) lambda*

alenahelenash

To solve the problem, we will set up the Lagrangian for the utility maximization subject to the budget constraint.
The Lagrangian is given by the formula:
$L\left({x}_{1},{x}_{2},\lambda \right)=U\left({x}_{1},{x}_{2}\right)-\lambda ·\left(2{x}_{1}+8{x}_{2}-80\right)$
where $U\left({x}_{1},{x}_{2}\right)$ is the utility function, $\lambda$ is the Lagrange multiplier, and $2{x}_{1}+8{x}_{2}-80$ represents the budget constraint.
Given that the utility function is $U\left({x}_{1},{x}_{2}\right)={x}_{1}^{\alpha }·{x}_{2}^{1-\alpha }$ and $\alpha =0.9$, we can substitute these values into the Lagrangian:
$L\left({x}_{1},{x}_{2},\lambda \right)={x}_{1}^{\alpha }·{x}_{2}^{1-\alpha }-\lambda ·\left(2{x}_{1}+8{x}_{2}-80\right)$
Now, we will differentiate the Lagrangian with respect to ${x}_{1}$, ${x}_{2}$, and $\lambda$ to find the critical points:
$\frac{\partial L}{\partial {x}_{1}}=\alpha ·{x}_{1}^{\alpha -1}·{x}_{2}^{1-\alpha }-2·\lambda =0$ (1)
$\frac{\partial L}{\partial {x}_{2}}=\left(1-\alpha \right)·{x}_{1}^{\alpha }·{x}_{2}^{-\alpha }-8·\lambda =0$ (2)
$\frac{\partial L}{\partial \lambda }=2{x}_{1}+8{x}_{2}-80=0$ (3)
We also have the budget constraint: $2{x}_{1}+8{x}_{2}=80$.
To solve this system of equations, we can eliminate $\lambda$ by rearranging equation (3) to express $\lambda$ in terms of ${x}_{1}$ and ${x}_{2}$:
$2{x}_{1}+8{x}_{2}=80⟹2{x}_{1}=80-8{x}_{2}⟹{x}_{1}=40-4{x}_{2}$
Substituting this value of ${x}_{1}$ into equations (1) and (2), we can solve for ${x}_{2}$:
$\alpha ·\left(40-4{x}_{2}{\right)}^{\alpha -1}·{x}_{2}^{1-\alpha }-2·\lambda =0$ (4)
$\left(1-\alpha \right)·\left(40-4{x}_{2}{\right)}^{\alpha }·{x}_{2}^{-\alpha }-8·\lambda =0$ (5)
Next, we can solve equations (4) and (5) simultaneously to find the values of ${x}_{2}$ and $\lambda$. Once we have those values, we can substitute them back into the budget constraint equation $2{x}_{1}+8{x}_{2}=80$ to find the corresponding value of ${x}_{1}$.

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