# The corner points of the feasible region determined by the system of linear constraints are $(0,0),(0,40),(20,40),(60,20),(60,0)$.The objective function is $Z=4x+3y.$ Compare the quantity in column A and column B: \begin{array}{1 1}column\; A &amp; column\; B\\maximum\;of \;z &325\end{array}

$\begin{array}{1 1}(A)\;The\;quantity\; in\;column\;A\;is\;greater\\(B)\;The\;quantity\; in\;column\;B\;is\;greater\\(C)\;The\; two\; quantites\; are\; equal\\(D)\; The\; relationship\; cannot\; be \;determined\; on\; the\; basis\; of\; the\; information\; supplied\end{array}$

Toolbox:
• 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 given corner points are $(0,0),(0,40),(20,40),(60,20),(60,0)$
The objective function is $Z=4x+3y$
For the points $(x,y)$ the objective function subject to $Z=4x+3y$
Step 2:
At $(0,0)$ the objective function $z=4x+3y\Rightarrow z=0$
At $(0,40)$ the objective function $z=4x+3y\Rightarrow Z=4\times 0+3\times 40=120$
At $(20,40)$ the objective function $z=4x+3y\Rightarrow 4\times 20+3\times 40=200$
At $(60,20)$ the objective function $z=4x+3y\Rightarrow 4\times 60+3\times 20=300$
At $(60,0)$ the objective function $z=4x+3y\Rightarrow 4\times 60+3\times 0=240$
Step 3:
Hence the maximum of $Z$ is 300
For column B the maximum is 325.
Clearly quantity of column B is greater.
Hence the correct option is $B$.