$\begin{array}{1 1} (A)\;Z=50x+60y \\(B)\;Z=200x+60y \\ (C)\;Z=20x+30y \\(D)\;Z=20x+60y \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
- If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R

Step 1:

Let $X$ be the resistors,$Y$ be the transistors and $Z$ be the capacitors produced by the electronic circuits of two types $A$ and $B$

The data in the given problem can be written as follows :

For Type A: $x=20,y=10,z=10$.The profit on type $A$ circuit is Rs.50

For Type B: $x=10,y=20,z=30$.The profit on type $B$ circuit is Rs.60

The maximum requirement is $x=200,y=120,z=150$

Step 2:

These units are bought to fulfill the maximum requirement of $x,y,z$ and to maximize the profit.

The mathematic formulation of the above problem is as follows :

Maximize $Z=50x+60y$ subject to $2x+y\leq 20,x+2y\leq 12,x+3y\leq 15,x\geq 0,y\geq 0$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...