Question

This Lagrange multiplier technique works because of two facts for any f(x, y) and constraint function k = g(x, y):

• ∇g is always perpendicular to the curve k = g(x, y),, where in this case g(x,y)=x^2/4+y^2

• the maximums and minimums of f(x, y) on k = g(x, y) occur when ∇f is perpendicular to the curve as well.

a.Plug the parameterization r(t)= (2cost,sint) of the ellipse into ∇g=(x/2,2y)  and verify that for all values of t it is perpendicular to r'(t).

b) Now plug r(t) into ∇f=(y,x) and verify that ∇f and  r'(t) are perpendicular precisely at the points (sqrt2,1/sqrt2), (-sqrt2,-1/sqrt2),(-sqrt2,1/sqrt2),(sqrt2,-1/sqrt2)

c)Explain why you think, in general, the maximum and minimum values of f(x, y) will occur at points where ∇f is perpendicular to the curve k = g(x, y)

Answer 1