USE THIS REFERENCE FOR EACH LETTER: a.) L = b.)

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:
$L(x_1,x_2,\lambda )=U(x_1,x_2)-\lambda ·(2x_1+8x_2-80)$
where $U(x_1,x_2)$ is the utility function, $\lambda$ is the Lagrange multiplier, and $2x_1+8x_2-80$ represents the budget constraint.
Given that the utility function is $U(x_1,x_2)=x_1^{\alpha }·x_2^{1-\alpha }$ and $\alpha =0.9$, we can substitute these values into the Lagrangian:
$L(x_1,x_2,\lambda )=x_1^{\alpha }·x_2^{1-\alpha }-\lambda ·(2x_1+8x_2-80)$
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)
$\frac{\partial L}{\partial x_2}=(1-\alpha )·x_1^{\alpha }·x_2^{-\alpha }-8·\lambda =0$ (2)
$\frac{\partial L}{\partial \lambda }=2x_1+8x_2-80=0$ (3)
We also have the budget constraint: $2x_1+8x_2=80$.
To solve this system of equations, we can eliminate $\lambda$ by rearranging equation (3) to express $\lambda$ in terms of $x_1$ and $x_2$:
$2x_1+8x_2=80⟹2x_1=80-8x_2⟹x_1=40-4x_2$
Substituting this value of $x_1$ into equations (1) and (2), we can solve for $x_2$:
$\alpha ·(40-4x_2{)}^{\alpha -1}·x_2^{1-\alpha }-2·\lambda =0$ (4)
$(1-\alpha )·(40-4x_2{)}^{\alpha }·x_2^{-\alpha }-8·\lambda =0$ (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 $2x_1+8x_2=80$ to find the corresponding value of $x_1$.