Question

ME In Andrew's Furniture Shop, he assembles both bookcases and TV stands. Each type of furniture takes him about the same time to assemble. He figures he has time to make at most 18 pieces of furniture by this Saturday. The materials for each bookcase cost him $20.00 and the materials for each TV stand cost him $40.00. He has $600.00 to spend on materials. Andrew makes a profit of $60.00 on each bookcase and a profit of $100.00 for each TV stand. Find how many of each piece of furniture Andrew should make so that he maximizes his profit. Using the information in the problem, write the constraints. Let a represent number of bookcases, and y represent number of TV stands.

needed: verticies, optimization equation, maxiumum profit and what he needs to sell to achieve that profit.​

Answer 1

Answer:6 bookcases, 12 TV stands.

Explanation:

We can model this using a system of equations. Let b represent how many bookcases he makes, and t how many TV stands he makes. He can make at most 18 pieces, so let's say that b+t <=18. The cost of the bookcases are 20, and the TV stands are 40, so 20b+40t gives us our total cost. We have 600 to spend, so 20b+40t <=600. We want to maximize the equation for profit, which is 60b+100t=P. We have three values we care about, where our inequalities intersect, if we have only TVs, and if we only have bookcases. But, if b=0 and t=18, then our cost is 720, which is too big. So if we only have tv stands, then we can have at most 15 (40*15=600). So our three cases are: (b=0, t=15), (t=0, b=18), and the intersection, which is where the inequalities are equal. To solve that, b=18-t, 20(18-t)+40t=600, 360-20t+40t=600, 360+20t=600, 20t=240, t=12. b=18-t, b=18-12, b=6. So our final case is (b=6, t=12). Now simply plug those three cases into our equation to maximize, 60*0+15*100=1500. 18*60+0*100=1080. 6*60+12*100=1560. So, our third case yields us the highest profit, and Andrew should make 6 bookcases and 12 TV stands.