A feasible region has vertices at (1,7), (2,4), and (3,6). What is the minimum value of C=3x+5y
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A feasible region has vertices at (1,7), (2,4), and (3,6). What is the minimum value of C=3x+5y

[From: ] [author: ] [Date: 11-12-15] [Hit: ]
Thanks!-Well, the graph z=3x+5y is a plane in three dimensions, coordinates (x,y,z).......
I just changed schools and I have an exam for Algebra II Honors. Please tell me HOW to o this, DO NOT just give me the answers. Also, i am new to this ahoo answers thingyy. Thanks!

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Well, the graph z=3x+5y is a plane in three dimensions, coordinates (x,y,z).

The feasible region is a polygon drawn on the x-y plane that tells you what region we are concerned with. We are looking for the minimum of z in this polygon.

The three vertices form a triangle in this case. The point (1,7) in the feasible region is the same as (1,7,0) in this three dimensional space. The point (1,7,C=38) is the point directly above the feasible region lying in the plane z=3x+5y.

Because z=3x+5y is a plane, maximums and minimums occur only at the vertices of the feasible region. (This is a key theorem to know).

Check by plugging in the vertices, finding the minimum.

Plug in (1,7) and I get a value C=38
Plug in (2,4) and I get a value C=26
Plug in (3,6) and I get a value of C=39

So, in this feasible region, the point (3,6) produces the maximum value for C, and at (2,4) a minimum. We dont have to concern ourselves with any point on the edges or any point inside the feasible region. Maximums and minimums only occur at vertices.
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