Hello Holly,
1.) We are given the objective function:
$$f(x,y)=x^2+y^2$$
subject to the constraint:
$$g(x,y)=x+2y-20=0$$
Lagrange multipliers gives rise then to the following system:
$$2x=\lambda(1)$$
$$2y=\lambda(2)$$
This implies:
$$\lambda=2x=y$$
Substituting for $y$ into the constraint, we find:
$$x+2(2x)-20=0\implies x=4\implies y=8$$
Thus, we obtain the critical point:
$$(x,y)=(4,8)$$
The objective function's value at this point is:
$$f(4,8)=4^2+8^2=80$$
To ensure this is a minimum, let's evaluate the objective function at another point on the constraint, such as $$(x,y)=(2,9)$$:
$$f(2,9)=2^2+9^2=85$$
And so we may now conclude:
$$f_{\min}=f(4,8)=80$$
2.) We are given the objective function:
$$f(x,y)=\sqrt{99-x^2-y^2}$$
subject to the constraint:
$$g(x,y)=x+y-10=0$$
Now here we see that $x$ and $y$ have cyclic symmetry, that is, we may switch the two variables and still have the same objective function and constraint. Thus we know the critical value may be obtained when $x=y$. The constraint then gives us:
$$x=y=5$$
And so our critical point is:
$$(x,y)=(5,5)$$
The objective function's value at this point is:
$$f(5,5)=\sqrt{99-5^2-5^2}=7$$
To ensure this is a maximum, let's evaluate the objective function at another point on the constraint, such as $$(x,y)=(4,6)$$:
$$f(5,5)=\sqrt{99-4^2-6^2}=\sqrt{47}<7$$
And so we may conclude:
$$f_{\max}=f(5,5)=7$$
