Economics maximization problem linear activity

Question

Consider the vectors:

$a_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \\0 \end{pmatrix}, a_2 = \begin{pmatrix} 0 \\ 0 \\ -1 \\1 \end{pmatrix}, a_3 = \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1\end{pmatrix}$

Find a single vector $p$ which maximizes $pa_i$ for $i = 1,2,3$.

To put this in context this is an economics profit max problem where p is a price and each component of the above vectors represents the quantity of the good.

I honestly have no idea how to find this $p$ vector. It doesn't even seem possible to me that a single vector can maximize these three vectors.

Answer 1

I think your question is incomplete. First you should determine your goal function, e.g., you can choose the sum of all profits to maximize, max: $pa_{1}+pa_{2}+pa_{3}$. Also you should determine your boundaries. E.g., total number of goods $d_1, d_2, d_3, d_3$.

Answer 2

A paraphrased version of this question is crossposted on Economics.SE, and as a result, we are closing it. Using the information available there and here, I believe I have an answer.

Basically, you can perform any activity any number of times. Suppose you perform $a_i$ $x_i$ times and the price of each good $j$ is $p_j$. Let $A$ denote the matrix defined by $a_1,a_2,a_3$, let $x$ denote the row vector $(x_1, x_2, x_3)$ of activites and let $p$ denote the column vector $(p_1, p_2, p_3)$ of prices. Then the problem is $$ \max_x x A p. $$ Suppose this maximum exists and $$ c^* = \max_x x A p, $$ $$ x^* = arg\max_x x A p. $$ As there is no constraint on $x$ the activity vector, $2 \cdot x^*$ yields a profit of $$ (2 \cdot x^*) A p = 2 \cdot c^*. $$ If $c^* > 0$, then $2 \cdot c^* > c^*$ so $c^*$ was not maximal after all. A similar argument exists for negative profits. Therefore if a profit maximum exists in case of linear activities without resource constraints it is zero. Hence you need to find a vector $p$ for which \begin{eqnarray*} a_1 \cdot p & = & 0 \\ \\ a_2 \cdot p & = & 0 \\ \\ a_3 \cdot p & = & 0. \end{eqnarray*} From here on out it is a straightforward exercise in linear algebra.