Answer to Question #176373 in Operations Research for kris

Question

1. An appliance dealer wants to purchase a combined total of no more than 100 refrigerators, and dishwashers for inventory. Refrigerators weigh 200 pound each, and dishwashers weigh 100 pounds each. The dealer is limited to a total of 12,000 pounds for these two items. A profit of $35 for each refrigerator and $20 on each dishwasher is projected.

(a) Write out the linear programming model by identifying the constraints and the objective function from the description above.

(b) Using a scale of 2 cm to 20 pounds on both axes, construct and shade the region R in which every point satisfies all the constraints.

(c) Based on the graph obtained in (b), determine the corner points and find out the maximum number of refrigerators and dishwashers that the dealer can purchase and sold to make the profit.

Answer 1

Solution:

(a): Let the number of refrigerators and dishwashers be $x$ and $y$ respectively.

Objective function, $Z=35x+20y$ to maximise and

subject to constraints:

$200x+100y\le12000\Rightarrow2x+y\le120 \\ x+y\le100 \\ x,y\ge0$

(b): Consider these inequations to be equations and plotting them on graph.

Put (0,0) in these inequations.

$0\le120\Rightarrow$ True

$0\le100\Rightarrow$ True

So, the shadow or bounded area of these inequations will be towards O(0,0) or including it.

Taking 2 cm to 20 pounds scale is too small for displaying this bounded area and inequations as they are large in numbers. So, we take each 50 units to 500 pounds, which is same as the r

equired one.

(c): Solving above equations $x+y=100, 2x+y=120$ to get intersection point $B(20,80)$

Clearly, OABC is a feasible area.

Corner points are $O(0,0),A(60,0),B(20,80),C(0,100)$

At $O(0,0), Z=35(0)+20(0)=0$

At $A(60,0), Z=35(60)+20(0)=2100$

At $B(20,80), Z=35(20)+20(80)=700+1600=2300$

At $C(0,100),Z=35(0)+20(100)=2000$

Clearly, 2300 is maximum when $x=20,y=80$

To get maximum profit, the number of refrigerators and dishwashers are 20 and 80 respectively.