Andrew should make 6 bookcase and 12 tv stand to maximize his profut $1560.
Given that 18 pieces of furniture is made and the materials for each bookcase cost him $20.00 and the materials for each TV stand cost him $40.00 and he has $600.00 to spend on materials. He makes a profit of $60.00 on each bookcase and a profit of $100.00 for each TV stand.
Let x represent the number of book case and let y represent number of tv stand.
Total number of book case and total number of tv stand is x+y≤18 ......(1)
Total amount of book case and total amount of tv stand is 20x+40y≤600 ......(2)
minimum amount of book case is x≥0 and minimum amount of tv stand is y≥0.
Now, solve the inequality (1) by converting it into equality and substituting x=0 and y=0, we get
x+y=18
When x=0 then
0+y=18
y=18
When y=0 then
x+0=18
x=18
Now, solve the inequality (2) by converting it into equality and substituting x=0 and y=0, we get
20x+40y=600
When x=0 then
20(0)+40y=600
40y=600
y=15
when y=0 then
20x+40(0)=600
20x=600
x=30
Further, we will grph the inequalities and we get feasible region and taking boundary point from graph.
From the graph the vertices of the feasible region is (0,0),(6,12),(0,15) and (18,0)
Further, we will find the maximization profit for the equation Z=60x+100y
When (x,y)=(0,0) then
Z=60(0)+100(0)
Z=0
When (x,y)=(6,12) then
Z=60(6)+12(100)
Z=360+1200
Z=1560
When (x,y)=(0,15) then
Z=60(0)+12(15)
Z=0+1500
Z=1500
When (x,y)=(18,0) then
Z=60(18)+100(0)
Z=1080
Hence, maximum profit is $1560 when andrew should build 6 bookcase and 12 tv stand.
Learn more about inequality from here brainly.com/question/9774970
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