Answer:
6 bookcases, 12 TV stands.
Step-by-step 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.
