Optimization 11-24-2020

Using $z^2=1-x^2-y^2$ , we have $\begin{aligned}x+2y+2z^2&=x+2y+2(1-x^2-y^2)\\ &=2+x-2x^2+2y-2y^2\\ &=2-2\left(x-\frac14\right)^2+\frac18-2\left(y-\frac12\right)^2+\frac12\\ &\leq2+\frac18+\frac12\\ &=\frac{21}8=2.625\end{aligned}$ This value of $\frac{21}8$ is attained at $x=\frac14$ and $y=\frac12$ , hence $z^2=1-\frac1{16}-\frac14=\frac{11}{16}$ , so $z=\pm\frac{\sqrt{11}}4$ . Hence, this maximum is attained at a legitimate point $(x,y,z)$ .

The constrained optimization problem can be converted to an unconstrained optimization problem by recognising that the optimal solution must lie on the unit sphere. This means:

$x = \sin{\theta} \cos{\phi} \ ; \ y = \sin{\theta} \sin{\phi} \ ; \ z = \cos{\theta}$

$0 \le \theta \le \pi \ ; \ 0 \le \phi \le 2 \pi$

This converts the cost function to:

$F = \sin{\theta} \cos{\phi} + 2 \sin{\theta} \sin{\phi} + 2\cos^2{\theta}$

Now, one can compute the partial derivative of $F$ with respect to $\theta$ and $\phi$ and proceed from there, but I have resorted to a shortcut. I swept across the parameter space using a short script of code which is attached below. Code is commented. The maximum value is $\boxed{\approx 2.625}$ .

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clear all
clc

% Numerical resoloution:
dphi   = pi/1000; dtheta = pi/1000;

% Variable counter:
c = 1;

% Nested loop to sweep over the the unit sphere:
for theta = 0:dtheta:pi
    for phi = 0:dphi:2*pi

        % Cost function evaluation:
        F(c)  = sin(theta)*cos(phi) + 2*sin(theta)*sin(phi) + 2*cos(theta)^2;
        c     = c + 1;
    end
end

% Printing maximum value:
max(F)

Substituting the spherical constraint in for $z^2$ gives us a function in two variables $f(x,y) = x + 2y + 2(1-x^2-y^2) = x+2y-2x^2-2y^2+2.$ Taking $grad f = 0$ yields:

$f_{x} = 1-4x = 0 \Rightarrow x =\frac{1}{4},$

$f_{y} = 2-4y=0 \Rightarrow y = \frac{1}{2}.$

The Hessian Matrix of $f$ computes to:

$F(x,y) = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} = -4 \cdot I_{2x2}$

which is negative-definite for all $x,y \in \mathbb{R}.$ Hence, the maximum value of $f$ is just $f(\frac{1}{4},\frac{1}{2}) = \boxed{2.625}.$