Question

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 1

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

$Z=2A+3B$

where A = no of toys of A and B = no of toys of B

Constraints are

$A+B\leq 2000\\8A+14B\leq 20000$

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 $(1333,667) = 2666+2001\\=4667$

Profit for $(2000,0) =4000$

Profit for $(0.1428)=4284$

Since maximum at (1333,667) we find that

to maximize profit A = 1333 and B = 667 to be produced.