A. see below for a graph
B. f(x, y) = f(0, 15) = 90 is the maximum point
A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)
B. For f(x, y) = 4x +6y, we have ...
f(0, 0) = 0
f(0, 15) = 6·15 = 90 . . . . . the maximum point
f(10, 0) = 4·10 = 40
Comment on evaluating the objective function
I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.
1. you have to describe a test so just tell of a test you took that was about graphing and has the symbols <, ≤, ≥, >
2. > or < is dashed line
≥ or ≤ is solid line
3. what do you think?
So you need to find y and multiple that by 55 and you will get answer
We should use a test point to determine the region or regions of the graph of an inequality should be shaded because by plotting an inequality the graph is divided into two regions : one which lie in feasible region and the other does not .
Now in order to check the feasible region we start by taking a test point and see whether it makes the inequality true or not.If it does not makes the inequality true then that region is not shaded and other is shaded.If it satisfy the inequality then the region containing that point is shaded.
Graph the inequality, use a test point to check.
y = 3x - 4...