A population gathers plants and animals for survival. They need at least 420 units of energy, 300 units of protein, and 8 hides. One unit of plants provides 30 units of energy, 10 units of protein, and 0 hides. One animal provides 20 units of energy, 25 units of protein, and 1 hide. Only 22 units of plants and 30 animals are available. It costs 20 hours of labor to gather one unit of a plant and 10 hours for an animal. Find how many units of plants and animals should be gathered to meet the requirements with a minimum number of hours of labor.

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Let x represent the number of plants gathered and y the number of animals.

The objective function is C=20x+10y (20 hours per unit of plants, and 10 hours per unit of animal) -- we wish to minimize the objective function.

We are subject to the following constraints:

`30x+20y>=420` or `3x+2y>=42` (The amount of energy taken in must be at least 420 units.)

`10x+25y>=300` or `2x+5y>=60` (The amount of protein must be at least 300 units.)

`0x+y>=8` or `y>=8` (The number of hides must be at least 8 -- so the number of animal units must be at least 8.)

`0<=x<=22` (The natural constraint of no negative plant units allowed, and the maximum number of plant units available is 22.)

`0<=y<=30` The number of animal units available is 30.

Here is a graph of the feasible region:

There are 6 points of intersection bounding the feasible region. They are (0,21),(0,30),(22,30),(22,8),(10,8),`(90/11,96/11)` .

C(0,21)=210

C(0,30)=300

C(22,30)=740

C(22,8)=520

C(10,8)=280

`C(90/11,96/11)=2760/11~~250.91`

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The minimum occurs when 21 animals are gathered and no plants.

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