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Determine vector x $\in R^3$ with $\|x\|^2=x^Tx=1$ which maximizes the function below

$$ f(x) = 2x_1^2 +2x_2^2-x_3^2+2x_1x_2$$

If someone can show me how to tackle this problem then I have at least a starting point or if you can tell me how to solve this then I can try it myself. Any help is appreciated !

Michael Hardy
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Elise
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  • You're maximizing inside the unit sphere in $\mathbb{R}^3$. Do the standard $\nabla f = 0$ to find extrema in the interior, then set the condition $x_1^2 + x_2^2 + x_3^2 = 1$ and optimize again. Compare solutions and identify the largest. – user217285 Jun 26 '15 at 22:36

2 Answers2

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Check Newton's method for optimization.

Basically, compute $x_1 = x_0 - \frac{f'(x)}{f''(x)}$.

You should find a analytical solution.

davcha
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Your limitation on $x=(x_1,x_2,x_3)$ is $x_1^2+x_2^2+x_3^2=1$. That gives us

$$f(x)=2x_1^2 +2x_2^2-x_3^2+2x_1x_2$$ $$=2(x_1^2+x_2^2+x_3^2)-3x_3^2+2x_1x_2$$ $$=2-3x_3^2+2x_1x_2$$

The limitations place $(x_1,x_2,x_3)$ on the surface of the unit sphere. Clearly we maximize $f(x)$ by making $x_3=0$ and maximizing $2x_1x_2$ for $x_1^2+x_2^2=1$.

Geometrically, that last is finding the point on the unit circle for which the hyperbola defined by $2x_1x_2=k$ has the largest value of $k$. A quick look at the graph shows that happens at $x_1=x_2=\frac{\sqrt{2}}2$.

enter image description here

So your desired vector is

$$x=\left(\frac{\sqrt 2}2,\frac{\sqrt 2}2,0 \right)$$

Rory Daulton
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