# A manufacturer produces two types of steel trunks. He has two machines A and B. The first type of trunk requires 5 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and 2 hours on machine B.Machines A and B can work at most for 24 hours and 15 hours per day, respecively. He earns a profit of Rs. 30 and RS. 25 per trunk on the first type and second type, respecively. How many trunk of each type must be made each day to make the maximum profit?

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 data given in the problem can be summarized as follows :
Let $x$ units of type A and y units of type B be produced to fulfill the requirement.
Step 2:
The mathematical formulation is as follows :
Maximize $Z=30x+25y$
Subjected to $3x+3y\leq 18$ and $3x+2y\leq 15$ and $x,y\geq 0$
Here $Z$ denotes the maximum profit.
Step 3:
To solve this LPP graphically let us convert the inequalities into equation and draw the corresponding lines.
$3x+3y=18$
$x+y=6$
$3x+2y=15$
Clearly OAPC is the feasible region which is shaded.
The corner points are $O(0,0),A(5,0),P(3,3),C(0,6)$
Step 4:
Now let us obtain the values of Z as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=30x+25y$
At $O(0,0)$ the value of the objective function $Z=0$
At $A(5,0)$ the value of the objective function $Z=30(5)+25(0)=150$
At $P(3,3)$ the value of the objective function $Z=30(3)+25(3)=165$
At $C(0,6)$ the value of the objective function $Z=30(0)+25(6)=150$
Clearly $Z$ has maximum value of 165 at the point $P(3,3)$
Hence 3 trunks of each type is required to be produced to obtain maximum profit.