Answer:
The maximum value of the objective function is P=500
Step-by-step explanation:
we have
[tex]x+y\leq 18[/tex] ----> constraint A
[tex]x+3y\leq 30[/tex] ----> constraint B
[tex]x\geq 0[/tex] ----> constraint C
[tex]y\geq 0[/tex] ----> constraint D
Solve the system of inequalities by graphing
The solution is the shaded area
see the attached figure
The vertices of the shaded area are
(0,0), (0,10),(12,6),(18,0)
Find the maximum value of the objective function
P=10x+50y
For each vertex substitute the value of x and the value of y in the objective function
1) For (0,0)
P=10(0)+50(0)=0
2) For (0,10)
P=10(0)+50(10)=500
3) For (12,6)
P=10(12)+50(6)=420
4) For (18,0)
P=10(18)+50(0)=180
therefore
The maximum value of the objective function is P=500
