Yefim's answer is nice and direct, as usual (although I think there is a small mistake). Lagrange multipliers is definitely how you're meant to do these problems, and you need to know it. But I wanted to show a second way that works because of the simple shape. It's fun because there are many shortcuts.

For interior critical points, change nothing. But for the boundary, it's the unit circle! So (x,y) = (cos θ, sin θ)! So

g(θ) = f(x(θ),y(θ)) = 4- cos^{2}(θ) - sin^{4}(θ) + (1/2) sin^{2}(θ) and then

g'(θ) = 2cos(θ)sin(θ) - 4sin^{3}(θ)cos(θ) + sin(θ)cos(θ)

= cos(θ) sin(θ) [3 - 4sin^{2}θ]

If we set that to 0 (i.e. all three factors to 0), we get solutions

θ = ±π/2, 0, π, ±π/3, ±2π/3 .

That looks like a lot of work! But since all the trig functions are to an even power, they won't give negative answers. So we can pretend all solutions are in the first quadrant (the others will be redundant). So

g(π/2) = f(0,1)

g(0) = f(1,0) (missing from above)

g(π/3) = f( 1/2, √3/2 ) (missing from above)

g(0) is the minimum.

Tom K.

08/14/20