**Section 11.8 Problem 7**: Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y,z)=xyz; x^2 +2y^2 +3z^2 =6

Comment: Noting that the constraint gives us a closed bounded region, the Lagrange multiplier equations would give us all critical points provided that the gradient of the left hand side of the constraint is not zero.

**Section 11.8 Problem 16**: Find the extreme values of f subject to both constraints. f(x,y,z)=x^2 +y^2 +z^2; x-y=1, y^2 -z^2 =1

Solution: We can express x=1+y and z^2=y^2-1. Then f(x,y,z)=(1+y)^2+y^2+y^2-1=3y^2+2y. Notice that y^2-1\geq 0, that is, y\leq -1 or y\geq 1. Easy to see 3y^2+2y reaches its minimum 1 when y=-1 and has no maximum.

**Section 11.8 Problem 19**: Find the extreme values of f on the region described by the inequality. f(x,y)=e^{-xy}, x^2 +4y^2 \leq 1

Comment: For the interior of the region x^2+4y^2<1, we use the partial derivatives to find the critical points. For the boundary, we can use Lagrange multiplier to find the extremes.

**Section 11.8 Problem 44**: The plane 4x - 3y + 8z = 5 intersects the cone z^2 = x^2 + y^2 in an ellipse.

- Graph the cone and the plane, and observe the resulting ellipse.
- Use Lagrange multipliers to find the highest and lowest points on the ellipse.

Comment for 2: The function we want to optimize is f(x,y,z)=z subject to the constraints 4x-3y+8z=5 and x^2+y^2-z^2=0. Noting that the gradients of the left hand sides can never be zero and the constraints give us a closed bounded region, we can use Lagrange multiplier equations to find the extremes.