question_answer

Solve the linear programming problem: max \[Z=x+2y\]subject to constraints: \[x-y\le 10,\]\[2x+3y\le 20.\],

A)  \[max\text{ }Z=10~\]

B)  \[max\text{ }Z=30\]

C)  \[max\text{ }2=40\]       

D)  \[max\text{ }Z=50\]

E)  None of these

Correct Answer: E

Solution :

We have, max \[Z=x+2y\] Subject to constraints, \[x-y\le 10\] \[2x+3y\le 20\] \[x\ge 0,\,\,y\ge 0\] On taking given constraints as equation, we get The following graph.   Here, OAB is the required feasible region whose comer points are \[O(0,\,\,0),\,\,\,A(10,\,\,0)\] and \[B\left( 0,\frac{20}{3} \right)\].

Comer point	\[z=x+2y\]
at   \[O(0,\,\,0)\]	\[Z=0\]
at \[A\,(10,\,\,0)\]	\[Z=10\]
at \[B\left( 0,\frac{20}{3} \right)\]	\[Z=0+2\times \frac{20}{3}=\frac{40}{3}\]

Hence, maximum value of Z is \[\frac{40}{3},\] which is obtained at \[B\left( 0,\frac{20}{3} \right).\]