Question

A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of lower quality. Profits on each type of belt are Rs. 2 and Rs. 1.50 per belt, respectively. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belts of type B, only 700 buckles are available per day. How should the company manufacture the two types of belts in order to have a maximum overall profit?

Answer 1

We assume the number of belts produced of type A is $x$ and type $B$ is $y$. Since the number of belts cannot be negative we have the non-negative constraint $x\ge 0,y\ge 0$. We use the given data of the maximum number of belts that can be produced per day is 800 we have the inequality constraint $x+y\le 800$ and from the data of a number of buckles we have the inequality constraints $x\le 400,y\le 700$. We have from the data of profit the optimizing profit function $P=2x+1.5y$. We solve the linear programming problem graphically to maximize the profit.

Complete step-by-step solution
Let us assume that the number of belts produced of type A is $x$ and type $B$ is $y$. Since the number of belts produced cannot be negative we have;
$x\ge 0,y\ge 0......\left( 1 \right)$
We are given that the number of belts that can be produced both A and B combined is at maximum 800 because of leather deficiency. So we have;
$x+y\le 800......\left( 2 \right)$
We are given that Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belts of type B, only 700 buckles are available per day. Since we can attach on buckle per one belt we have;

& x\le 400.....\left( 3 \right) \\\ & y\le 700.....\left( 4 \right) \\\ \end{aligned}$$ We are given profits on each type of belt A is Rs. 2 and type of belt is B Rs. 1.50 per belt, respectively which means $x$ number of belts are produced with profit Rs. 2 and $y$ number of belts are produced with profit function. So we have the profit function as; $$P=2x+1.5y....\left( 5 \right)$$ If we put $\left( 0,0 \right)$ in constraints (2) , (3) and (4) they are going to satisfy. So the feasible region contains the origin and we see from constraint (1) that the feasible region will be in the first quadrant in the first quadrant. We convert equations (1), (2), (3), and (4) into equalities to have; $$x=0,y=0,x+y=800,x=400,y=700$$ We plot this line on graph paper and use the inequality constraints to find the feasible region OABCD as shaded below.  We know that optimal value (here maximum) occurs at one of the vertices of the feasible region which here are $O\left( 0,0 \right),A\left( 400,0 \right), B\left( 400,400 \right), C\left( 100,700 \right), D\left( 0,700 \right)$. We find the cost function in the below table. Points $\left( x,y \right)$| Profit (in rupees)$P=2x+1.5y$ ---|--- O(0,0)| 0 A(400,0)| 800 B(400,400)| 1400 C(100,700)| 1250 D(0,700)| 1050 We see that the maximum profit occurs at $B\left( 400,400 \right)$ for 1400 rupees per day. So the company needs to produce 400 bets of type A and type B to maximize the profit.$$$$ **Note:** We can find the coordinates of the vertices $A, B, C, D$ by solving the equalities of the constraints taking two at a time. We can alternatively the optimal value by moving the line $2x+1.5y=k$ (called profit line) towards the origin for any $k$ and the optimal point will be the point which meets the profit line first. A constraint of rate production may also be added to the question.