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October 2, 2014

October 2, 2014

Posted by **Carl** on Monday, December 10, 2012 at 6:40pm.

x<(or equal too)1 y>(or equal too)0

3x + y<(or equal too)5

2. Describe the linear programming situation for this system of inequalities where you are asked to find the maximum value of f(x, y) = x + y.

x>(or equal too) 0 y>(or equal too)0

6x + 3y<(or equal too)18 x + 3y<(or equal too)9

- Pre Calc -
**Damon**, Monday, December 10, 2012 at 6:52pmleft (or at) of x = 1

above or on x axis

below or on y = -3x+5

which hits x axis at x = 5/3

and hits the y axis at y = 5

sketch that

You see that the vertical line x = 1 hits our sloped line somewhere above the x axis

find that point

x = 1

y = -3(1) + 5 = 2

so

we have a corner at (1, 2)

everything in the upper half plane left of the sloping line and the vertical line x = 1

It did not say x had to be positive, so it goes left forever. That would not be likely in a real linear programming problem.

- Pre Calc -
**Damon**, Monday, December 10, 2012 at 7:12pm2. Describe the linear programming situation for this system of inequalities where you are asked to find the maximum value of f(x, y) = x + y.

x>(or equal too) 0 y>(or equal too)0

6x + 3y<(or equal too)18 x + 3y<(or equal too)9

=================================

In first quadrant due to x>/=0 and y>/= 0

6x + 3y<(or equal too)18

is

y </= -2x + 6

x and y axis intercepts at x = 3 and at y = 6

below that line

3 y </= -x + 9

y </= -(1/3) x + 3

intercepts at x = 9 and at y = 3

below that line

we need the corner where those sloped lines hit

-(1/3) x + 3 = -2x+6

-x + 9 = -6 x + 18

5 x = 9

x = 9/5

then y = -(1/3)(9/5) + 3

= 3 - 3/5 = 12/5

So we have a corner at (9/5 , 12/5)

so three corners to test

(0 , 3) , (9/5 , 12/5) , (3 , 0)

at (0,3) x+y = 3

at (9/5,12/5) x+y = 21/5 = 4. something

at (3,0) x+y = 3 again

so the max is at the intersection of the sloped lines and is 21/5

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