Simultaneous Inequalities

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May 13th 2011, 06:37 AM
mathguy80

Simultaneous Inequalities

Hi Guys,

This one is long problem. I figured out most of it, but need a hint with the last proof part.

Draw a diagram illustrating the region S of the xy-plane which is defined by the simultaneous inequalities $x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19$ , and give the coordinates of the vertices of S. Prove that, if line $y = kx$ intersects S, then $\dfrac{1}{6} \le k \le \dfrac{5}{2}$ .

The Point P lies on y = kx and is in the region S. Prove that, when $\dfrac{1}{6} \le k \le \dfrac{3}{5}$ , the maximum value for the y-coordinate of P is $\dfrac{13k}{2 + k}$ and find the corresponding expression when $\dfrac{3}{5} \le k \le \dfrac{5}{2}$

I completed the initial parts of the problem. Got the 3 vertices of the region S, $A = (6,1), B=(5,3), C = (2,5)$

Then since y = kx must pass through A or C, got max and min slopes and hence proved, $\dfrac{1}{6} \le k \le \dfrac{5}{2}$ .

I am stuck at the next part of the problem. Any ideas? Thanks for your help!
May 14th 2011, 04:45 AM
Sudharaka

Quote:

Originally Posted by mathguy80

Hi Guys,

This one is long problem. I figured out most of it, but need a hint with the last proof part.

Draw a diagram illustrating the region S of the xy-plane which is defined by the simultaneous inequalities $x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19$ , and give the coordinates of the vertices of S. Prove that, if line $y = kx$ intersects S, then $\dfrac{1}{6} \le k \le \dfrac{5}{2}$ .

The Point P lies on y = kx and is in the region S. Prove that, when $\dfrac{1}{6} \le k \le \dfrac{3}{5}$ , the maximum value for the y-coordinate of P is $\dfrac{13k}{2 + k}$ and find the corresponding expression when $\dfrac{3}{5} \le k \le \dfrac{5}{2}$

I completed the initial parts of the problem. Got the 3 vertices of the region S, $A = (6,1), B=(5,3), C = (2,5)$

Then since y = kx must pass through A or C, got max and min slopes and hence proved, $\dfrac{1}{6} \le k \le \dfrac{5}{2}$ .

I am stuck at the next part of the problem. Any ideas? Thanks for your help!

Dear mathguy80,

Since, $\dfrac{1}{6} \le k \le \dfrac{3}{5}$ , the line lies in between the points A and B.

Consider the line which goes through A and B. Its equation is,

$\frac{y-1}{x-6}=\frac{1-3}{6-5}=-2$

$y=-2x+13$

Now the point P lies on the line $y=kx$ .

Since k>0, it is clear that y would be a maximum when the point P lies on AB (Please refer to your diagram).

Therefore, $kx=-2x+13$

$x=\frac{13}{k+2}$

Therefore, $y=\frac{13k}{k+2}$

Hope this will help you.
May 25th 2011, 03:54 AM
mathguy80

Hey @Sudharaka. I was away for a while and forgot about this thread! Your solution checks out. Thanks and sorry for the late reply.
May 25th 2011, 04:56 AM
Sudharaka

Quote:

Originally Posted by mathguy80

Hey @Sudharaka. I was away for a while and forgot about this thread! Your solution checks out. Thanks and sorry for the late reply.

You are welcome.