Graph the feasible region for the follow system of inequalities by drawing a polygon around the feasible region.
2x+6y≤54
7x+4y≤70
x≥0
y≥0
Feasible regions and polygons, oh my! Let's see what kind of shape we can whip up!
To start, let's graph each inequality individually and shade in the solutions.
For the first inequality, 2x + 6y ≤ 54, let's rewrite it in slope-intercept form, shall we? So, y ≤ -(1/3)x + 9. To make life easier, we'll assume x = 0 and y = 0 (because who likes math overcomplicating things?). Plotting those points on a graph, we get a nice little line sloping downwards. Shade the area beneath the line, my friend.
Now, for the second inequality, 7x + 4y ≤ 70, rewriting it in slope-intercept form gives us y ≤ -(7/4)x + 17. Once again, we'll assign x = 0 and y = 0, and plot those points. This line has a negative slope, so it goes downward at an angle. Shade in the region below this line as well.
So far, so good! Now let's tackle the last two inequalities, x ≥ 0 and y ≥ 0. These bad boys are just telling us to stay positive, so let's keep the graph in the first quadrant (where x and y values are greater than or equal to 0).
Alright, now it's time to connect the dots! Well, not really dots, more like the shaded regions. Draw lines to connect the vertices of the shaded regions together and voilà! You've got yourself a polygon-shaped feasible region.
Put on your party hats, because we've just created some math art with our polygon. Now you can identify all the x and y values that fall within that polygon - those are your feasible solutions. Enjoy!