Answer:
The maximum profit is $1600 at x=30 and y=10.
Step-by-step explanation:
Let x be the number of units of product X.
y be the number of units of product Y.
The profit for X is $35 and the profit for Y is $55.
Maximize [tex]Z = 35x + 55y[/tex] ..... (1)
It requires 4 pounds of material and 2 hours of labor to produce one unit of X. It requires 4 pounds of material and 6 units of labor to produce one unit of Y.
Total material = 4x+4y
Total labor = 2x+6y
There are 160 pounds of material and 120 hours of labor available.
[tex]4x+4y\leq 160[/tex] .... (2)
[tex]2x+6y\leq 120[/tex] ..... (3)
[tex]x\geq 0,y\geq 0[/tex]
The related line of inequality (2) and (3) are solid line because the sign of equality "≤" contains all the point on line in the solution set.
Check the inequalities by (0,0).
[tex]4(0)+4(0)\leq 160[/tex]
[tex]0\leq 160[/tex]
This statement is true.
[tex]2x+6y\leq 120[/tex]
[tex]2(0)+6(0)\leq 120[/tex]
[tex]0\leq 120[/tex]
It means shaded region of both inequalities contain (0,0).
The extreme points of common shaded region are (0,0), (0,20), (40,0) and (30,10).
At (0,0),
[tex]Z = 35(0) + 55(0)=0[/tex]
At (0,20),
[tex]Z = 35(0) + 55(20)=110[/tex]
At (40,0),
[tex]Z = 35(40) + 55(0)=140[/tex]
At (30,10),
[tex]Z = 35(30) + 55(10)=1600[/tex]
Therefore the maximum profit is $1600 at x=30 and y=10.
