Question

A store sells two types of toys, A and B. The store owner pays Rs. 8 and Rs. 14 for each unit of toy A and toy B respectively. One unit of toy A yields a profit of Rs. 2 while a unit of toy B yields a profit of Rs. 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 Rs. 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

Let x be the total number of toys A and y the number of toys B.

x and y cannot be negative. Hence,

$x\ge 0$ and $y\ge 0$

The store owner estimates that no more than 2000 toys will be sold every month.

$x+y\le 2000$

One unit of toy A yields a profit of Rs. 2 while a unit of toy B yields a profit of Rs. 3, hence the total profit P is given by

$P=2x+3y$

The store owner pays Rs. 8 and Rs. 14 for each one unit of toy A and B respectively and he does not plan to invest more than Rs. 20,000 in inventory of these toys.

$8x+14y\le 20000$

The solution set of the system of inequalities given above and the vertices of the region obtained are shown in the figure.

Vertices of the solution set :

A at (0,0)

B at (0,1429)

C at (1333,667)

D at (2000,0)

Calculate the total profit P at each vertex.

P(A) = 0

P(B) = 4287

P(C) = 4667

P(D) = 4000

The maximum profit is at vertex C with x = 1333 and y = 667.

Hence, the store owner has to have 1333 toys of type A and 667 toys of type B in order to maximize his profit.