What point in the feasible region maximizes the objective function, 3x + y ≤ 12, x+y ≤5, x ≥0,y ≥0

Step-by-step explanation:

We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.

Steps to graph 3x + y ≤ 12:

First we graph 3x + y = 12 then shade the graph for ≤.

plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.

plug x=0

3x + y = 12

3(0) + y = 12

0 + y = 12

y = 12

Hence first point is (0,12)

Similarly plugging x=2 will give y=6

Hence second point is (2,6)

Now graph both points and joint them by a straight line.

test for shading.

plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:

3(0) + (0) ≤ 12

0 + 0 ≤ 12

0 ≤ 12

Which is true so shading will be in the direction of test point (0,0)

We can repeat same procedure to graph other inequalities.

From graph we see that ABCD is feasible region whose corner points will result into maximum or  minimu for objective function.

Since objective function is not given in the question so i will explain the process.

To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer

answer: 48

just remember to use pemdas:

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