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Given a vector $\vec{c}$ and a radius $r$, solve the problem:

\begin{equation*} \begin{aligned} & \underset{x}{\text{maximise}} & & \vec{c} \cdotp \vec{x}=a \\ & \text{subject to} & & ||x^2||=r^2 \end{aligned} \end{equation*}

jp412
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1 Answers1

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Let $\theta$ be the angle between $\vec{x}$ and $\vec{c}$. Then assuming the constraint is $\|\vec{x}\|^2 = r^2$, we have

$$ \vec{c} \cdot \vec{x} = \|\vec{c}\| \|\vec{x}\| \cos\theta = \|\vec{c}\| |r| \cos\theta $$

which is maximised when $\theta = 0$.

The geometric interpretation is that for $\vec{x}$ of fixed length $|r|$, the dot product $\vec{c} \cdot \vec{x}$ is maximised by taking $\vec{x}$ pointing in the same direction as $\vec{c}$,

$$ \vec{x} = |r| \frac{\vec{c}}{\|\vec{c}\|} .$$

  • Thanks for the response, however, I was wondering how to approach this if $\vec{c}$ and $\vec{x}$ were $n$x$1$ vectors. Sorry this wasn't clear in the question. Am I correct in assuming the dot product relationship only holds for 3-D space? – jp412 Apr 20 '16 at 07:22
  • @jp412 $\vec{c} \cdot \vec{x} = |\vec{c}| |\vec{x}| \cos\theta$ works in $\mathbb{R}^n$ for all positive integer $n$, with $\theta$ the angle between the vectors. –  Apr 20 '16 at 07:28
  • Okay thanks for your help – jp412 Apr 20 '16 at 07:35