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:

$L({x}_{1},{x}_{2},\lambda )=U({x}_{1},{x}_{2})-\lambda \xb7(2{x}_{1}+8{x}_{2}-80)$

where $U({x}_{1},{x}_{2})$ 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 $U({x}_{1},{x}_{2})={x}_{1}^{\alpha}\xb7{x}_{2}^{1-\alpha}$ and $\alpha =0.9$, we can substitute these values into the Lagrangian:

$L({x}_{1},{x}_{2},\lambda )={x}_{1}^{\alpha}\xb7{x}_{2}^{1-\alpha}-\lambda \xb7(2{x}_{1}+8{x}_{2}-80)$

Now, we will differentiate the Lagrangian with respect to ${x}_{1}$, ${x}_{2}$, and $\lambda $ to find the critical points:

$\frac{\partial L}{\partial {x}_{1}}=\alpha \xb7{x}_{1}^{\alpha -1}\xb7{x}_{2}^{1-\alpha}-2\xb7\lambda =0$ (1)

$\frac{\partial L}{\partial {x}_{2}}=(1-\alpha )\xb7{x}_{1}^{\alpha}\xb7{x}_{2}^{-\alpha}-8\xb7\lambda =0$ (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 ${x}_{2}$:

$2{x}_{1}+8{x}_{2}=80\u27f92{x}_{1}=80-8{x}_{2}\u27f9{x}_{1}=40-4{x}_{2}$

Substituting this value of ${x}_{1}$ into equations (1) and (2), we can solve for ${x}_{2}$:

$\alpha \xb7(40-4{x}_{2}{)}^{\alpha -1}\xb7{x}_{2}^{1-\alpha}-2\xb7\lambda =0$ (4)

$(1-\alpha )\xb7(40-4{x}_{2}{)}^{\alpha}\xb7{x}_{2}^{-\alpha}-8\xb7\lambda =0$ (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 ${x}_{1}$.