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**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 $\displaystyle x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19$, and give the coordinates of the vertices of S. Prove that, if line $\displaystyle y = kx$ intersects S, then $\displaystyle \dfrac{1}{6} \le k \le \dfrac{5}{2}$.

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

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

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

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