Finding Corners Without The Graph | Free Math Help Forum

• Home
• Forums New posts Search forums
• What's new New posts Latest activity

Log in Register What's new Search

Search

Everywhere Threads This forum This thread Search titles only By: Search Advanced search…

• New posts
• Search forums

Menu Log in Register Install the app Install

• Forums
• Free Math Help
• Advanced Math

You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Finding corners without the graph

• Thread starter juicy123
• Start date Jan 11, 2006

J

juicy123

New member

Joined Jan 3, 2006 Messages 13 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.

stapel

Super Moderator

Staff member Joined Feb 4, 2004 Messages 16,550

juicy123 said: Finding corners without the graph Click to expand...

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. J

juicy123

New member

Joined Jan 3, 2006 Messages 13 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

stapel

Super Moderator

Staff member Joined Feb 4, 2004 Messages 16,550 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. You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

• Forums
• Free Math Help
• Advanced Math

• This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register. By continuing to use this site, you are consenting to our use of cookies. Accept Learn more…

Top