Given that <em>f(x, y)</em> is a linear function, and the constraints themselves are also linear, it follows that the vertices of the feasible region are the sites of extrema of <em>f(x, y)</em>. So just find where each boundary line intersects with another line, and check the value of <em>f(x, y)</em> at each intersection.
We have
• <em>x</em> = 0 and <em>y</em> = 0 ===> (0, 0)
• <em>x</em> = 0 and 2<em>x</em> + 7<em>y</em> = 70 ===> <em>y</em> = 10 ===> (0, 10)
• <em>y</em> = 0 and 8<em>x</em> + 4<em>y</em> = 136 ===> <em>x</em> = 17 ===> (17, 0)
• 2<em>x</em> + 7<em>y</em> = 70 and 8<em>x</em> + 4<em>y</em> = 136 ===> (14, 6)
At these points, we respectively get
• <em>f</em> (0, 0) = 0
• <em>f</em> (0, 10) = 60
• <em>f</em> (17, 0) = 34
• <em>f</em> (14, 6) = 64
Then max <em>f(x)</em> = 64 at (14, 6) and min <em>f(x)</em> = 0 at (0, 0).
