Finding corners without the graph

juicy123

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Jan 3, 2006
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When a profit function is linear and the feasible region is a polygon, the profit function will always achieve its maximum at a corner point of the feasible region. But for problems involving three variables, drawing the feasible region can be difficult. (And it’s impossible for more than three variables!)

So it’s helpful to be able to locate the corner points without actually drawing out the region. As preparation for more complex cases, consider the two-variable feasible region defined by these linear inequalities.

X+ 2y < 8
_
(x plus 2y lesser or equal to 8)

2x+y < 13
_
(2x plus y lesser or equal to 13)

y< 3
_
(y lesser than 3)

x> 0
_
(x greater than 0)

y>0
_
(y is greater than 0)

1. Each of theses inequalities has a corresponding linear equation, whose graph is a straight line, and each corner point of the feasible region is the intersection of two of these lines. How many combinations of these equations are there, taking them two at a time?
2. For each of your combinations in question 1, fine the intersection pint of the pair of lines. (If theses if no intersection point, explain why not.)
3. Which of the intersection points from question 2 are actually corner points of the feasible region defined by the inequalities? Explain how you know.
 
juicy123 said:
Finding corners without the graph
Work with the associated equalities, and pair them off. Then solve the "systems" that you've created.

For instance:

. . . . .x + 2y < 8
. . . . .2x + y < 13

...becomes:

. . . . .x + 2y = 8
. . . . .2x + y = 13

...which is solved by whatever method you like, such as:

. . . . .-2x - 4y = -16
. . . . . .2x + y = 13

. . . . .-3y = -3
. . . . .y = 1

. . . . .x + 2(1) = 8
. . . . .x + 2 = 8
. . . . .x = 6

So the two lines cross at (x, y) = (6, 1).

Form all the pairs, solve all the systems, and then test the optimization equation at each "corner".

Note: In "real life", there are specialized processes (and software) that are used to do the more complex optimization problems.

Eliz.
 
okay, so you use substitution for # 2 and 3... that's what i thought but i wasnt sure.

i still dont get what #1 is asking for? or how to get there
 
1) My interpretation is that they're asking how many pairs of lines you can get out of the listed lines. That is, how many systems of equations are you going to have to solve?

So list the equations. The first one will be paired, one at a time, with each of the others. How many pairs does that make? The second one has already been paired with the first one; this leaves the third through last equations. How many additional pairs does this make?

And so forth.

Eliz.
 
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