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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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In the figure below ,the feasible region (shaded)for a LPP is shown .Determine the maximum and minimum value of $Z=x+2y$

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  • 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:
The corner points of the feasible regions are $P(\large\frac{3}{13},\frac{24}{13})$,$Q(\large\frac{3}{2},\frac{15}{4})$,$R(\large\frac{7}{2},\frac{3}{4})$ and $S(\large\frac{18}{7},\frac{2}{7})$
The function with in this constraint is $Z=x+2y$
Now let us find the maximum and minimum values of the function ,with in the given constraint.
Step 2:
The values of the objective function at these points are :
For the point $P(\large\frac{3}{13},\frac{24}{13})$ the value of the objective function is $Z=x+2y\Rightarrow \large\frac{3}{13}$$+2\times \large\frac{24}{13}=\frac{51}{13}$
For the point $P(\large\frac{3}{2},\frac{15}{4})$ the value of the objective function is $Z=x+2y\Rightarrow \large\frac{3}{2}$$+2\times \large\frac{15}{4}=$$9$
For the point $R(\large\frac{7}{2},\frac{3}{4})$ the value of the objective function is $Z=x+2y\Rightarrow \large\frac{7}{2}$$+2\times \large\frac{3}{4}=$$5$
For the point $S(\large\frac{18}{7},\frac{2}{7})$ the value of the objective function is $Z=x+2y\Rightarrow \large\frac{18}{7}$$+2\times \large\frac{2}{7}=\frac{22}{7}$
Step 3:
The maximum value is $9$ at $Q(\large\frac{3}{2},\frac{15}{4})$
The minimum value is $\large\frac{22}{7}$ at $S(\large\frac{18}{7},\frac{2}{7})$
answered Aug 20, 2013 by sreemathi.v
 

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