Question

A dealer wishes to purchase toys A and B. He has Rs.580 and has space to store 40 items. A costs Rs.75 and B costs Rs.90. He can make profit of Rs.10 and Rs.15 by selling A and B respectively. Assuming that he can sell all the items that he can buy formulation of this as L.P.P. is
$
A. Maximize{\text{ }}z = 10x + 15y \\
  Subject{\text{ }}to{\text{ }}x + y \leqslant 40,\,75x + 90y \leqslant 580,x \geqslant 0,\,y \geqslant 0 \\
$
$
B. Maximize{\text{ }}z = 10x + 15y \\
  Subject{\text{ }}to{\text{ }}x + y \geqslant 40,\,,x \geqslant 0,\,y \geqslant 0,75x + 90y \geqslant 580 \\
$
$
C. Maximize{\text{ }}z = 15x + 10y \\
  Subject{\text{ }}to{\text{ }}x + y \leqslant 40,\,75x + 90y \leqslant 580,x \geqslant 0,\,y \geqslant 0 \\
$
$
D. Maximize{\text{ }}z = 10x + 15y \\
  Subject{\text{ }}to{\text{ }}x + y \geqslant 40,\,75x + 90y \leqslant 580,x \geqslant 0,\,y \geqslant 0 \\
$

Answer 1

Hint: We can assume x as the number of toy A and y as the number of the toy B. Our aim is to get maximum profit. We can form a function of profit from the profit per toy of each type. Then the sum of costs must be less than or equal to the money he has. The total number of toys must be less than or equal to the storage space and the number of each toy must be greater than or equal to 0. We can form these constraints and check with the options to get the correct option.

Complete step-by-step answer:
Let x be the number of toys A and y be the number of toys B.
It is given that the profit by selling A and B respectively are Rs.10 and Rs.15. We need to maximize the profit. Let z be the function of profit. Then z is given by,
$z = 10x + 15y$
It is given that he has a space to store only 40 items. Therefore, the total number of toys must be less than or equal to 40.
$ \Rightarrow x + y \leqslant 40$
It is given that toy A costs Rs.75 and B costs Rs.90. It is also given that he has only Rs.580. So, the total cost must be less than or equal to 580. So, we can write the inequality as,
$ \Rightarrow 75x + 90y \leqslant 580$
As x and y are numbers of toys, they cannot be negative.
$ \Rightarrow x \geqslant 0,\,y \geqslant 0$
Therefore the function we need to maximize is $z = 10x + 15y$ and the constraints are $x + y \leqslant 40,\,75x + 90y \leqslant 580,x \geqslant 0,\,y \geqslant 0$
So, the correct answer is option A.

Note: LPP or linear programing problems are used to find the maximum or minimum of a function subjecting to specified constraints. It is also called optimization problems. For formulating an LPP, firstly, we need to find the function that has to be minimised or minimized. Then we must find all constraints. To find the maximum or minimum value, we plot the inequalities in a graph to get a polygon. Then we find the values of the function at the vertices. The point at which the value of the function is maximum will be maximise the function.