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The solution to self-quiz 49 of math 210, which involves finding the maximum and minimum values of a function and identifying critical points using lagrange multipliers and the second derivative test. The problem includes finding the maximum and minimum values of x + 3y for points on an ellipse and analyzing the behavior of the function x² + y³ − 2y².
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Self-quiz 49
(a) Compute fx, fy, fxx, fxy and fyy.
(b) Find all critical points.
(c) Classify each critical point as a maximum, minimum or saddle.
Self-quiz 49
Solution: We will be using the method of Lagrange multipliers. To this end, we set:
f (x, y) = x + 3y g(x, y) = x^2 + 4y^2 − 13
and we consider the system:
{ ∇f = λ∇g g = 0
which can be rewritten as:
1 = 2λx 3 = 8λy x^2 + 4y^2 = 13
We solve for x and y to get:
x =
2 λ y =
8 λ
and then plug into the third equation to get:
4 λ^2
64 λ^2
that yields:
1 = 16λ^2 =⇒ λ = ±
To study the behaviour of the function at those points we will compute the determinant for the second derivative test. In general this determinant equals:
∣∣ fxx^ fxy fxy fyy
0 12 y^2 − 4
∣∣ = 24y^2 − 8
So at (0, 0) we get D = − 8 < 0 which means that this is a saddle point. At (0, 1) we get D = 16 which means that we have a local extremum. In particular, since fxx(0, 1) = 2 > 0, this has to be a local minimum. Finally, at (0, −1) we get D = 16 and fxx(0, −1) which means that we have another local minimum.
