5 numbers whose sum is 1 and their squares sum is 11, maximum value of their cubes

Question

What will be the maximum of the sum of cubes of 5 numbers where the sum of these numbers is 1 and the sum of the squares of these numbers is 11. I am guessing the answer is 25 (3, -1, -1, 0, 0 being the numbers). But how do we find these using Lagrangian multipliers?

Answer 1

The constraint functions are $$g_1(x) = \sum_{i=1}^5 x_i - 1$$ $$g_2(x) = \sum_{i=1}^5 x_i^2 - 11.$$

The function to maximize is

$$f(x) = \sum_{i=1}^5 x_i^3.$$

The Lagrange conditions $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$ read (component-wise)

$$3x_k^2 = 2\lambda_1x_k + \lambda_2, \text{ for } k = 1,2,3,4,5$$ $$g_j(x)=0, \text{ for } j = 1,2$$

Solving these is a bit tedious, but we get

$$\lambda_1 = \frac{3}{20}(4+9\sqrt 6)$$ $$\lambda_2 = \frac{3}{50}(106-9\sqrt 6)$$

and one of the components of $x$ is $$\frac{1+6\sqrt 6}{5}$$ and others are $$\frac{1-3\sqrt {\frac{3}{2}}}{5}$$

This gives the maximum value of $f$

$$\frac{163+243\sqrt 6}{25} \approx 30.329$$