Finite Math Question

Question

A store sells two types of toys, A and B. The store owner pays $8 and $10 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 2200 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

You can solve this with Linear Programming or with Algebra.  I'll use Algebra.  Let x = the number of toy A.  The number of toy B will be 2200-x.

 

Cost = $8x + $10(2200-x)

$20,000 ≥ 8x + 22,000 - 10x   (we want our cost to be no more than $20,000)

-2000 ≥ -2x

1000 ≤ x  (switched the inequality sign around because I divided by -2)

So I must buy at least 1000 of toy A.

 

Profit = $2x + $3(2200-x)

Profit = 2x + 6600 -3x

Profit = 6600 - x

 

The cost equation tells me that I must buy at least 1000 of toy A (x ≥ 1000).  The larger that x is, the smaller the profit (since we are subtracting x from $6600).  Thus the max profit occurs when x = 1000.  Buy 1000 of toy A and 2200-1000 = 1200 of toy B.  Profit will be $6600 - $1000 = $5600.