Question

A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item of A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item A and 8 minutes for B . The profit per item of \(A\) is \(Rs 25\) and per item of \(B\) is ₹ 18 . The formulation of an L.P.P. to maximize the profit (where \(x\) is number of items A and \(y\) is the number of item \(B\) ) is

• A. \(\begin{array}{cc} & \text{ Maximize }\mathrm{z}=25x+18y \\ & \text{ subject to }20x+15y\leq 640 \\ & 5x+8y\geq 500 \\ & x,y\geq 0 \end{array}\)
• B. Maximize \(z=25x+18y\) \(\begin{array}{cc} \text{ subject to }20x+15y & \leq 640 \\ 5x+8y & \leq 500 \\ x,y & \geq 0 \end{array}\)
• C. \(\begin{array}{c} \text{ Maximize }z=25x+18y \\ \text{ subject to }20x+5y\leq 8 \\ 5x+8y\leq 10 \\ x,y\geq 0 \end{array}\)
• D. \(\begin{array}{cc} & \text{ Maximize }\mathrm{z}=25x+18y \\ & \text{ subject to }4x+3y\leq 128 \\ & \begin{array}{c} 5x+8y\geq 500 \\ x,y\geq 0 \end{array} \end{array}\)

Hint:

Convert the machine hours to minutes and write one inequality per machine with the processing times.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(x\) items of A and \(y\) items of B be made. Each machine gives a time constraint, and the profit is the objective function.

Step 2: Machine I
Available time: 10 h 40 min \(=640\) minutes. Time used: \(20x+15y\), so \(20x+15y\leq640\).

Step 3: Machine II
Available time: 8 h 20 min \(=500\) minutes. Time used: \(5x+8y\), so \(5x+8y\leq500\).

Step 4: Objective
Maximize \(z=25x+18y\) with \(x,y\geq0\).

Final Answer:
Maximize profit with both machine constraints of the less-than-or-equal type. \[ \boxed{\text{(B) }\text{Maximize }z=25x+18y\text{ subject to }20x+15y\leq640,\ 5x+8y\leq500,\ x,y\geq0} \]