Answer to Question #172072 in Operations Research for ONG SIU TING

Question

A catering company is to make lunch for a business meeting. It will serve salami sandwiches, and vegetarian sandwiches. A salami sandwich has 1 serving of vegetables, 4 slices of salami, 1 slice of cheese, and 2 slices of bread. A vegetarian sandwich has 3 servings of vegetables, 2 slices of cheese, and 2 slices of bread. A total of 10 bags of salami are available, each of which has 40 slices; 18 loaves of bread are available, each with 14 slices; 200 servings of vegetables are available, and 15 bags of cheese, each with 60 slices, are available. Given the resources, how many of each sandwich can be produced if the goal is to maximize the number of sandwiches? Please show all the steps of the solution

Answer 1

Solution:

Let the number of salami sandwiches be $x$ and that of vegetable sandwiches be $y$.

Objective function: $Z=x+y$

subject to the constraints:

$x+3y\le200 \\ 4x\le400 \Rightarrow x\le100 \\ x+2y\le900 \\ 2x+2y\le252 \Rightarrow x+y\le126$

Consider them equations and plot on graph, then taking the bounded region by inequalities, we get following graph:

Consider origin as O. Then, OABCD is the feasible region.

Solving the above equations to get their intersection points as follows:

$O(0,0),A(0,\dfrac{200}3),B(89,37),C(100,26),D(100,0)$

At $O,\ Z=0+0=0$

At $A,\ Z=0+\dfrac{200}3=\dfrac{200}3$

At $B,\ Z=89+37=126$

At $C,\ Z=100+26=126$

At $D,\ Z=100+0=100$

Maximum occurs at B and C, i.e., 126.

We find that maximum occurs at line segment joining B(89, 37) and C(100,26).

Thus, either 89 salami sandwiches and 37 vegetable sandwiches should be product or 100, 26 respectively or any point satisfying line segment BC.