plot the contraint areas 2x+3y=30
Now we have a nice theorem that states the max, and min, will occur on the boundry where constraint lines meet. So test the profit function 20x+30y at each corner, and you will find the max.
let's look at each region.
2x + 3y > 30
the boundary is 2x + 3y = 30
when x = 0, y = 10
when y = 0 , x = 15 , so we have (0,10) and (15,0)
plot these points on the axes and draw a dotted line,
shade in the region above this line
2x+y < 26
repeat my method, shade in the region below the line 2x+y = 26
careful with the last one, I will do that one as well.
-6x+y < 50 -----> y < 6x+50
draw the boundary line y = 6x+50
when x=0, y = 50 , when y = 0 x = -25/3
so your two points are (0,50) , (-25/3,0)
shade in the region below that line
notice that P = 20x + 30y has the same slope as 2x + 3y = 30
so the farthest point to the right of your region will be the solution.
Solve the corresponding equations to find that point, plug into
P = 20x + 30y
Which of the following coordinates are intercepts of the linear relation 2x- 3y + 30 = 0 ? I. (0 , 10) II.(0 , 2/3) III.(-10 , 0) IV.(-15, 0) A. I only B. I and IV only C. II and III only D. II and IV only This was what I was
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