$\begin{array}{1 1} (A)\;z=100x+170y \\(B)\;z=170x+100y \\ (C)\;z=3x+4y \\ (D)\;z=4x+3y \end{array} $

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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 data in the given problem can be summarized as follows :

A box of type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine.

A box of type B screws requires 8 minutes on the threading machine and 2 minutes on the slotting machine.

In a week ,each machine is available for 60hours(60$\times$ 60minutes=3600 minutes)

The company gets a profit of Rs100 per box on type A screws and Rs170 per box on type B screws.

Step 2:

Let $x$ be the no of screws required to be on a slotting machine and $y$ be the no of screws required to be on a threading machine.

These screws are brought to fulfill the maximum requirement of $x,y$ and maximize the profit.

Step 3:

The mathematic formulation of the above problem is as follows :

Maximize $z=10x+170y$

Subject to $3x+2y\leq 3600$ and $x+4y\leq 1800,x\geq 0,y\geq 0$

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