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By graphing the systems of constraints, find the values of x and y that maximize the objective function.

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

maximum for N = 100x+40y 

answers:

(0,0)

(5,0)

(0,8)

(2,6)

 Oct 25, 2015

Best Answer 

 #2
avatar+118608 
+10

PayPay sent me a private message asking if i could try to explain some more.  

Paypay doesn't understand the moving orange line.  

It is a strange concept I think, I am not surprised that you did not understand and I am pleased that you asked about it. laugh

 

 

I have changed the graph just a little. (I just added the names of the points)

https://www.desmos.com/calculator/4w3xa3c3yy

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

Now the dark quadrilateral in the middle is the region that x and y must lie in.

The vertices of this region are  (0,0),     (5,0),     (0,8),      (2,6)

Lets just substitute these points into the equation  100x+40y = constant

(0,0)       100*0+40*0=0                           The constant would be 0

(5,0)         100*5+40*0=500                     The constant would be 500

(0,8)        100*0+40*8=320                      The constant would be 320

(2,6)        100*2+40*6=200+240=440    The constant would b e 440

 

So the biggest constant is 500.   It happens whe x=5 and y=0     (5,0)

So you can do this without understanding the orange line but it would be good if you understood the line so I will try to explain.

 

 

Now, what has this got to do with the moving orange graph???

 

The orange line is the graph of  100x+40y = a constant 

What you are finding is the biggest constant that  this equation can equal (within the given region.)

The slider is N

So you can make N bigger or smaller.   I have set it to move between 0 an 1000.

So you can move the orange line, by changingthe value of N.   You want the biggest possible N value so that at least some part of the line falls in the given region. When N is 500 only one point of the line is in the region.  That point is (5,0).   If N is bigger than 500 then no point on the line will fall in the restricted region.  SO the biggest possible value of N within the given restrictions is 500 and N is 500 at the point (5,0)

 

Have a good think about this because it is a concept that can help you with a variety of problems.  :)

 Oct 25, 2015
 #1
avatar+118608 
+15

Hi PayPay      laugh

 

By graphing the systems of constraints, find the values of x and y that maximize the objective function.

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

Here is your graph.   smiley  

https://www.desmos.com/calculator/lmiz3pkgw2

 

The constraints for a quadrilateral where the vertices are the 4 points that you have been given

 

I have also graphed 100x+40y=N  where N can take on a sliding range of values.

It is orange and if you move the slider the orange line will move.

The greatest value of N will occur on the last corner befor the orange line leaves the constraint region.  Hopefully you can see that happen at (5,0)

maximum for N = 100x+40y 

 

Alternatively, if you substitute those answer values into the N equation, you will find that N is greatest at (5,0)

 

answers:

(0,0)

(5,0)

(0,8)

(2,6)

 Oct 25, 2015
 #2
avatar+118608 
+10
Best Answer

PayPay sent me a private message asking if i could try to explain some more.  

Paypay doesn't understand the moving orange line.  

It is a strange concept I think, I am not surprised that you did not understand and I am pleased that you asked about it. laugh

 

 

I have changed the graph just a little. (I just added the names of the points)

https://www.desmos.com/calculator/4w3xa3c3yy

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

Now the dark quadrilateral in the middle is the region that x and y must lie in.

The vertices of this region are  (0,0),     (5,0),     (0,8),      (2,6)

Lets just substitute these points into the equation  100x+40y = constant

(0,0)       100*0+40*0=0                           The constant would be 0

(5,0)         100*5+40*0=500                     The constant would be 500

(0,8)        100*0+40*8=320                      The constant would be 320

(2,6)        100*2+40*6=200+240=440    The constant would b e 440

 

So the biggest constant is 500.   It happens whe x=5 and y=0     (5,0)

So you can do this without understanding the orange line but it would be good if you understood the line so I will try to explain.

 

 

Now, what has this got to do with the moving orange graph???

 

The orange line is the graph of  100x+40y = a constant 

What you are finding is the biggest constant that  this equation can equal (within the given region.)

The slider is N

So you can make N bigger or smaller.   I have set it to move between 0 an 1000.

So you can move the orange line, by changingthe value of N.   You want the biggest possible N value so that at least some part of the line falls in the given region. When N is 500 only one point of the line is in the region.  That point is (5,0).   If N is bigger than 500 then no point on the line will fall in the restricted region.  SO the biggest possible value of N within the given restrictions is 500 and N is 500 at the point (5,0)

 

Have a good think about this because it is a concept that can help you with a variety of problems.  :)

Melody Oct 25, 2015

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