Minimization problem

Parametrizing the unit sphere with polar coordinates, so $(x,y,z) \; =\; (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ we want to minimize $x + 2y + 2z^2 \; = \; \sin\theta(\cos\phi + 2\sin\phi) + 2\cos^2\theta \; = \; \sin\theta(\cos\phi + 2\sin\phi) + 2 - 2\sin^2\theta$ over $0 \le \theta \le \pi$ , $0 \le \phi \le 2\pi$ . It is clear (since $\sin\theta \ge 0$ ) that $\sin\theta(\cos\phi + 2\sin\phi) + 2 - 2\sin^2\theta \; \ge \; 2 - \sqrt{5}\sin\theta - 2\sin^2\theta$ Since the function $f(x) = 2 - \sqrt{5}x - 2x^2$ is decreasing for $0 \le x \le 1$ , we deduce finally that $\sin\theta(\cos\phi + 2\sin\phi) + 2 - 2\sin^2\theta \; \ge \; 2 - \sqrt{5}\sin\theta - 2\sin^2\theta \; \ge \; 2 - \sqrt{5} - 2 \; = \; -\sqrt{5}$ Note that the value of $-\sqrt{5}$ is achieved when $\phi = \pi+\tan^{-1}2$ and $\theta = \tfrac12\pi$ . This makes the answer $\boxed{5}$ .

This problem can be solved with Lagrange mulitpliers. Introduce Lagrange Multiplier L. The equations then become:

1. 1 = 2 L x
2. 2 = 2 L y
3. 4 z = 2 L*z
4. x^2 + y^2 + z^2 = 1

If z is non-zero, equation 3 implies L = 2. Hence, x = 1/4 and y = 1/2 (from equations 1 and 2 respectively). Hence, equation 4 implies z = 1 or z = -1. Thus, the possible critical points if z is non-zero are (1/4, 1/2, 1) and (1/4, 1/2, -1).

If z = 0, equation 4 means x^2 + y^2 = 1. If x = 0, this means y is 1 or -1. Hence, two other possible critical points are (0, 1, 0) and (0, -1, 0).

Similarly, if z = 0 and y = 0, x is one or negative 1. Thus, other critical points are (1, 0, 0) and (-1, 0, 0)

Lastly, if z = 0 and both x and y are non-zero, we find from equations 1 and 2 that L = 1/(2 x) = (1/y). Thus, y = 2 x. Therefore, equation 3 implies:

5 x^2 = 1 or x = 1/sqrt(5) or x = -1/sqrt(5). Since y = 2 x in this case, we find critical points:

(1/sqrt(5), 2/sqrt(5), 0) and (-1/sqrt(5), -2/sqrt(5), 0).

Now, evaluating f at each of the critical points shows that the minimum value of f is -sqrt(5) so that a = 5,