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How do I set up this problem ?

A product can be made in three sizes, large, medium, and small, which yield a net unit profit of $12, 10$ and $9$ respectively. The company has three centers where this product can be manufactured and these centers have a capacity of turning out $550, 750$ and $275$ units of the product per day, respectively, regardless of the size or combination of sizes involved.

Manufacturing this product requires cooling water and each unit of large, medium, and small sizes produced require $21, 17$ and $9$ gallons of water, respectively. The centers $1, 2$ and $3$ have $10000,\;\, 7000$ and $4200$ gallons of cooling water available per day, respectively. Market studies indicate that there is a market for $700, 900$ and $340$ units of the large, medium, and small sizes, respectively, per day. By company policy, the fraction (scheduled production)/(center's capacity) must be the same at all the centers. How many units of each of the sizes should be produced at the various centers in order to maximize the profit?

Here is my idea :

decision variable $x_1$:the number of large product in center $1$

$x_2$:the number of medium product in center $1$

$x_3$:the number of small product in center $1$

$y_1$:the number of large product in center $2$

$y_2$:the number of medium product in center $2$

$y_3$:the number of small product in center $2$

$z_1$:the number of large product in center $3$

$z_2$:the number of medium product in center $3$

$z_3$:the number of small product in center $3$

Objection function : max:$12\times(x_1+y_1+z_1)+10\times (x_2+y_2+z_2)+9\times(x_3+y_3+z_3)$

constraint: $$21 x_1+17 x_2+9 x_3\le 10000$$

$$21 y_1+17 y_2+9 y_3\le7000$$

$$21 z_1+17 z_2+9 z_3\le4200$$

How do figure out the fraction (scheduled production)/(center's capacity) must be the same at all the centers? Thanks.

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$\frac {\text {scheduled production}}{\text {capacity}}= \frac {x_1+x_2+x_3}{550}=\frac {y_1+y_2+y_3}{750}=\frac {z_1+z_2+z_3}{275}$

Wanting these all to be the same fraction means you have two equations:

$\frac {x_1+x_2+x_3}{550}=\frac {y_1+y_2+y_3}{750}$ and $\frac {y_1+y_2+y_3}{750}=\frac {z_1+z_2+z_3}{275}$ (any two pairings will do; the third must follow).

The equations can be rewritten as:

$750x_1+750x_2+750x_3-550y_1-550y_2-550y_3=0$ and $275y_1+275y_2+275y_3-750z_1-750z_2-750z_3=0$

By the way, you also need to have the constraints:

$x_1+x_2+x_3\le 550$

$y_1+y_2+y_3\le 750$

$z_1+z_2+z_3\le 275$

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