Question 45411
Find the maximum value of  p=5x+2y under the following conditions.
x>0 y>0 2x+y<10 x+y<7

Draw an x,y coordinate system.
x>0 and y>0 mean we are looking at the 1st quadrant.

Graph y<-2x+10
Graph  y< -x+7
Find the coordinates of the four intersection points, as follows:
Find the intersection of the two lines, y=-2x+10 and y=-x+7; it is (3,4)
Find the intersection of y=-2x+10 and y=0; it is (5,0)
Find the intersection of y=-x+7 and x=0; it is (0,7)
Find the intersection of x=0 and y=0; it is (0,0)

Hopefully you can see these intersection points.

Now, evaluate p=5x+2y with the values of these intersection points, as follows:
(0,0); p=5(0)+2(0)=0
(0,7): p=5(0)+2(7)=14
(5,0): p=5(5)+2(0)=25
(3,4): p=5(3)2(4)=23

From this you can see that the maximum value of p is p=25 and 
that occurs when x=5 and y=0

Cheers,
Stan H.