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.