Yefim S. answered • 08/13/20
Math Tutor with Experience
Let find critical points inside disk x2 + y2 < 1;
∂f/dx = - 2x = 0; ∂f/dy = - 4y3 + y = 0; y = 0; y = ± 1/2.
We get 3 critical points (0, 0), (0, 1/2), (0, - 1/2), all inside disk. We have f(0,0) = 4; f(0, 1/2) = 4.0625; f(0, - 1/2) = 4.0625.
Now let concider board of disk x2 + y2 = 1.
Now we create Lagrange function: F(x,y = 4 - x2 - y4 + 1/2y2+ μ(x2 + y2 - 1);
∂F/∂x = - 2x + 2μx = 0, from this equation x = 0 or μ = 1
∂F/∂y = - 4y3 + y + 2μy = 0. If μ = 1 then - 4y3 + 3y = 0, y = 0, y = ±√3/2 but these points not belong to board of disk
∂F/∂μ = x2 + y2 - 1 = 0. If x =0 then y = ± 1; we get 2 points (0, 1) and (0, - 1)
f(0, 1) = 4 - 1 + 1/2= 3.5; f(0, -1) = 4 - 1 + 1/2 = 3.5.
So absolute maximum is f(0,± 1/2) = 4.0625 and absolute minimum is f(0, ± 1) = 3.5
Tom K.
Yefim has a great answer as always. Note that, as all variables are squared throughout, we could let u = x^2 and v = y^2 and consider f(u,v) = 4 - u - v^2 + 1/2 v on u + v <= 1, u, v >= 0. Given that df/du = -1, we note that all maxima will be on u = 0 and all minima on u + v = 1. For u = 0, we substitute and take the derivative with respect to v and thus find the maximum at v = 1/4 (note that the second derivative is negative). On u + v = 1, we can substitute for one variable in terms of the other. Remember to check the endpoints as well.08/14/20