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 $z=10x+15y$
  Subject to $x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$
• B. Maximize $z=10x+15y$
  Subject to $x+y\ge 40,x\ge 0,y\ge 0,75x+90y\ge 580$
• C. Maximize $z=15x+10y$
  Subject to $x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$
• D. Maximize $z=10x+15y$
  Subject to $x+y\ge 40,75x+90y\le 580,x\ge 0,y\ge 0$

Answer 1

The correct option is A Maximize $z=10x+15y$

Subject to $x+y\le 40,75x+90y\le 580,x\ge 0,y\ge 0$
Let the dealer have $x$ items of $A$ and $y$ items of $B$.

Clearly, $x\ge 0,y\ge 0$

The total number of items must not exceed $40$, and so $x+y\le 40$

Also, since he has only Rs.$580$ with him, the total cost must not exceed that.

$\Rightarrow 75x+90y\le 580$

Since we need to maximize the profit, which can be written as $z=10x+15y$since by selling $1$ item of $A$, the profit earned is Rs. $10$ and Rs. $15$ for $B.$