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Cornerpoints of solution region of linear system.

I am having trouble finding the cornerpoints of the solution region of a linear system. There seems to be discrepancy between the graph of the solution region and the cornerpoints I can find algebraically.

Here is the linear system:

x, y ≥0

1x + 11y ≤ 1

7x + 2y ≤1

This is a picture of the linear system plotted on a graph and of the system's small bounded solution region:

Attachment 26836

The cornerpoints that I have obtained algebraically are:

(i) for the intersection of y = 0 and x = 0, (0, 0)

(ii) for the intersection of y = 0 and 7x + 2y = 1, (1/7, -1/2)

(iii) for the intersection of x = 0 and 1x + 11y = 1, (1,0)

(iv) for the intersection of 1x + 11y = 1 and 7x + 2y = 1, (8.3, -0.7)

Those cornerpoints don't seem to be right according to the bounded region in the graph.

Here is an example of how I algebraically came to a cornerpoint:

Intersection of 1x + 11y = 1 and 7x + 2y = 1

(1-7x)/2 = (1-1x)/11

11(1-7x) = 2(1-1x)

11 - 77x = 2 - 2x

11 - 77x - 2 + 2x = 0

9 - 75x = 0

-75x = -9

x = 8.333333333

1(8.333333333) + 11y = 1

8.333333333 + 11y = 1

11y = -7.333333333

y = -0.666666666 or -0.7

Re: Cornerpoints of solution region of linear system.

Hi Yoodle15! :)

Quote:

Originally Posted by

**Yoodle15** -75x = -9

x = 8.333333333

Could you calculate that again?

Re: Cornerpoints of solution region of linear system.

$\displaystyle \frac{9}{75}= \frac{3(3)}{3(25)}= \frac{3}{25}= \frac{3(4)}{25(4)}= \frac{12}{100}$

Re: Cornerpoints of solution region of linear system.

Such a silly mistake on my part! Thank you for pointing it out to me! *feels silly*

Re: Cornerpoints of solution region of linear system.