A store sells two types of toys, A and B. The store owner pays $8 and $14 for each one unit of toy A and B respectively. One unit of toys A yields a profit of $2 while a unit of toys B yields a profit of $3. The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than $20,000 in inventory of these toys. How many units of each type of toys should be stocked in order to maximize his monthly total profit?

Answer:A = 1333 and B = 667 Step-by-step explanation:Given that a store sells two types of toys, A and B.       A B cost       8 14 Profit 2 3 Objective is to maximize profit [tex]Z=2A+3B[/tex]where A = no of toys of A and B = no of toys of BConstraints are[tex]A+B\leq 2000\\8A+14B\leq 20000[/tex]By solving this equation, we get(A,B)= (1333,667)Corner points would be lowest of x and y intercepts in both(A,B) = (2000,0) and (0,1428)Profit for [tex](1333,667) = 2666+2001\\=4667[/tex]Profit for [tex](2000,0) =4000[/tex]Profit for [tex](0.1428)=4284[/tex]Since maximum at (1333,667) we find thatto maximize profit A = 1333 and B = 667 to be produced.