Where did the other 7 equations go?

Consider linear functions $f,g,F$ on $\mathbb{R}^9$ defined by $f(\mathbf{x}) \; = \; \sum_{j=1}^9 x_j \hspace{1cm} g(\mathbf{x}) \; = \; \sum_{j=1}^9 jx_j \hspace{1cm} F(\mathbf{x}) \; = \; 100f(\mathbf{x}) - \sum_{j=1}^9 (10-j)^2x_j$ We are interested in finding the extreme values of $F$ over the domain $P$ , where $P \; = \; \big\{ \mathbf{x} \in \mathbb{R}^9 \; : \; f(\mathbf{x}) = 10\;,\;g(\mathbf{x}) \,= \, 18\;,\;x_1,x_2,x_3,...,x_9 \ge 0 \big\}$ Since $F$ is continuous and $P$ is compact, $F$ is bounded and achieves its bounds on $P$ . Let the maximum and minimum values of $F$ over $P$ be $M$ and $m$ respectively.

If $\mathbf{x} \in P$ then $x_2 + 2x_3 + 3x_4 + \cdots + 8x_9 \,=\, 18-10=8$ , and so $8+x_1 \; =\; x_1 + x_2 + 2x_3 + \cdots +8x_9 \; \ge \; f(\mathbf{x}) \; = \; 10 \hspace{1cm} 8 + 8x_1 \; = \; 8x_1 + x_2 + 2x_3 + \cdots + 8x_9 \; \le \; 8f(\mathbf{x}) \; = \; 80$ Thus $2 \le x_1 \le 9$ .

If $\mathbf{x} \in P$ and there exist $2 \le u < v \le 9$ such that $x_u,x_v > 0$ , consider the vector $\mathbf{y} \in \mathbb{R}^9$ defined by $y_j \; = \; \left\{ \begin{array}{lll} v-u & \hspace{1cm} & j = 1 \\ 1 - v & & j = u \\ u - 1 & & j = v \\ 0 & & \mathrm{o.w.} \end{array} \right.$ Then $f(\mathbf{y}) = g(\mathbf{y}) = 0$ , while $F(\mathbf{y}) = -(v-u)(u-1)(v-1) < 0$ . Since $x_1,x_u,x_v$ are all positive, it follows that there exists $\delta > 0$ such that $\mathbf{x} + \lambda\mathbf{y}$ belongs to $P$ for $|\lambda| < \delta$ . Since $F(\mathbf{x} + \lambda\mathbf{y}) \,=\, F(\mathbf{x}) + \lambda F(\mathbf{y})$ , we deduce that $F(\mathbf{x} + \lambda\mathbf{y}) < F(\mathbf{x})$ for all $0 < \lambda < \delta$ , while $F(\mathbf{x} + \lambda\mathbf{y}) > F(\mathbf{x})$ for all $-\delta < \lambda < 0$ . Thus $F(\mathbf{x})$ can be neither $m$ nor $M$ .

It is not possible for any $\mathbf{x} \in P$ to have $x_1$ as the only nonzero component. Thus, if $\mathbf{x} \in P$ is such that $F(\mathbf{x})$ is either $m$ or $M$ , then there must exist some $2 \le u \le 9$ such that $x_u > 0$ , but that $x_1$ and $x_u$ are the only nonzero components of $\mathbf{x}$ . This means that $x_1 \; = \; 10 - \tfrac{8}{u-1} \hspace{2cm} x_u \; = \; \tfrac{8}{u-1}$ and so $F(\mathbf{x}) \; = \; 100 \times 10 - \left\{9^2\left(10 - \tfrac{8}{u-1}\right) + (10-u)^2 \tfrac{8}{u-1}\right\} \; = \; 342 - 8u$ after some simplification. Thus $m$ must be the smallest of these values, and $M$ must be the largest. Hence $m \; = \; 342 - 8\times9 \; = \; 270 \hspace{2cm} M \; = \; 342 - 8\times2 \; = \; 326$ and so the sum of $m$ and $M$ is $270 + 326 = \boxed{596}$ .