a) find the possible area for x and y (this part is ok)
x+2y ≤ 300
x+y ≤ 175
3x + y ≤ 450
x ≥ 0
y ≥ 0
b) Find the highest value for z = 300x + 150
when x and y is in the possible area (I need help with this part)
+130511 We have five constraints....two of them, x ≥ 0 and y ≥ 0, just indicate that our answer - if any - will be found in the first quadrant
Here's a graph of the other three : https://www.desmos.com/calculator/tvty6v1szo
It can be shown that the objective function z = 300x + 150 is maximized at some corner point(s) where these three inequalities intersect......here, we only have one corner point intersection....and that occurs at (120, 90)
So.....zmax = 300(120) + 150 = 36,150
