How do I find any local maxima/minima of f(x, y, z) = (x^2)yz on the intersection of the planes x + y + z = 40 and x + y − z = 0 using the Lagrange multiplier method?

Answer 1

Let f(x,y,z) = x²yz g(x,y,z) =x+y+z =40 and h(x,y,z) =x+y -z =0. So ∇f =λ∇g +μ∇h. This gives 2xyz= λ+ μ , x²z= λ + μ, x²y =λ -μ. From this 2xyz = x²z and x=2y using this in the constraint equations gives 3y+z=40 and 3y-z =0 which gives 6y= 40 and y=20/3 and x=40/3. Using these in the constraint equation gives 60/3-z=0 and z=20. So the minima is (40/3,20/3,20).