Find the maximum value of the function f(x, y, z)= x +4y +5z on the curve of intersection of the plane x y + z =1 and the cylinder x2+ y2=1. SOLUTION We maximize the function f(x, y, z)= x +4y +5z subject to the constraints g(x, y, z)= x y + z =1 and h(x, y, z)= x2+ y2=1. The Lagrange condition is f =g +h, so we solve the equations (1)1=+2x(2)4=+2y(3)=(4)x y + z=1(5)x2+ y2=1. Putting = [from (3)] in (1), we get 2x=, so x =2/. Similarly, (2) gives y =9/(2). Substitution in (5) then gives 42+ =1 and so 2=, =97/2. Then x = , y = , and, from (4), z =1 x + y =1 . The corresponding values of f are 497+4997+511397=5 . Therefore the maximum value of f on the given curve is