1
$\begingroup$

Let $B=\{(x_1,...,x_5)\in\mathbb{R}^5|\sum_{i=1}^5x_i=6,\sum_{i=1}^5x_i^2=8\}$

Find $\max_{x\in B} x_5$.

My work:

I used Lagrange theorem to obtain this equation system:

$$\begin{cases} 0=\alpha+2\beta x_1 \\ 0=\alpha+2\beta x_2 \\ 0=\alpha+2\beta x_3 \\ 0=\alpha+2\beta x_4 \\ 1=\alpha+2\beta x_5 \end{cases}$$

The first time I summed all the equations and obtained: $1=5\alpha+12\beta$

The second time I multiplied the $i^{th}$ equation by $x_i$ and then summed up the equation which helps us obtain: $x_5=6\alpha+16\beta$

Now I'm stuck trying to find $\alpha,\beta $ to help me find $x_5$. Can anyone help?

  • 1
    What simplex? Or are you requiring $x_i \ge 0$? – 2017-01-29
  • 0
    You're right, it's not a simplex, I'll fix the title – 2017-01-29

1 Answers 1

1

You can't have $\beta = 0$, so you can solve each equation to get $x_i$ in terms of $\alpha$ and $\beta$. Then plugging these in to the constraints gives you two equations in $\alpha$ and $\beta$. Simplify and solve.