Calculus 3 Lagrange

Question

Find the maximum and minimum values of the function f(x,y)=x^2*y subject to 5x^2+4y^2=60

Answer 1

The lagrangian is

l(x,y) = x^2y + l(5x^2 + 4y^2 - 60)

You calculate lx and ly (the derivatives with respect to x and y) and set both equal to 0.

This will result in you determining the relationship between x and y at the critical point.

You then plug back into 5x^2 + 4y^2 = 60 to determine the values of x and y.

Note, just thinking about the problem, you realize that you will have two maxima (y positive, x positive and negative) and two minima (y negative, x positive and negative), as you can swap x and -x and get the same value for x^2y and 5x^2 + 4y^2, while swapping y and -y gives the same value for 5x^2 + 4y^2 but the negative of the value for x^2y. If you think about the function you are maximizing/minimizing on the ellipse, you also realize that you will have a local minimum and maximum at x = 0, but that will not be the global minimum/maximum.