“The Possibility” Restaurant
About the case
Angela Fox and Zooey Caulfield were close friends and nutrition majors. They decided to
open a French restaurant as there were no French restaurants at that location. They named
their restaurant “The Possibility”. They decided to go with two menus fish and beef.
Constraints




They would sell a maximum of 60 meals each night.
The time required for preparing fish was 15 minutes and 30 minutes for beef and a
total of 20 hrs of kitchen hour per day.
They would sell at least 3 fish dinners for every 2-beef dinners.
They also believe that at least 10% of their customers will order beef dinners.
In case there are 3 situations



We have to formulate LP model for Angela and Zooey to estimate the number of
meals and maximize profit.
If Angela and Zooey increased the menu price on the fish dinner so that the profit for
both dinners was the same, what effect would that have on their solution?
Suppose Angela and Zooey reconsidered the demand for beef dinners and decided
that at least 20% of their customers would purchase beef dinners. What effect would
this have on their meal preparation plan?
Solution Model
Let,
x1= No of fish meal
x2= No of beef meal
Objective function
Estimated profit of $12 from each fish dinner and $16 from each beef dinner
Max Z= 12x1+16x2
Constraints
x1+x2 ≤ 60………………………(i)
0.25x1+0.5x2 ≤ 20………………(ii)
𝑥1 3
≥
𝑥2 2
2x1-3x2 ≥ 0……………………..(iii)
𝑥2
≥ 0.1
𝑥1 + 𝑥2
-01.x1+0.9x2 ≥ 0………………(iv)
x1
x2
x1+x2<=60
0
60
0.25x1+0.5x2<=20
x1
0
80
x2
40
0
60
0
Solving graphically,
A
B Optimal Point
C
O
Solving graphically,
Point O= 0,0
Profit is Rs 0
Point A = 34.28,22.85
Max Z= 12x1+16x2
Putting the values;
Z= 776.96
Max Profit is $ 776.96
Point B= 40,20
Max Z= 12x1+16x2
Putting the values;
Z= 800
Max Profit is $ 800
Point C= 54,6
Max Z= 12x1+16x2
Putting the values;
Z= 744
Max Profit is $ 744
From the graphical solution, the optimal point is at B (40,20) and the maximum profit is $
800. They need to prepare 40 fish meals and 20 beef meals every night to make a maximum
profit of $ 800
Part 2
If Angela and Zooey increased the menu price on the fish dinner so that the profit for both
dinners was the same
New Objective function
Max Z= 16x1+16x2
All other constraints remain the same
x1+x2 ≤ 60………………………(i)
0.25x1+0.5x2 ≤ 20……………(ii)
𝑥1 3
≥
𝑥2 2
2x1-3x2 ≥ 0……………………..(iii)
𝑥2
≥ 0.1
𝑥1 + 𝑥2
-01.x1+0.9x2 ≥ 0………………(iv)
Point A = 34.28,22.85
Max Z= 16x1+16x2
Putting the values;
Z= 914.08
Max Profit is $ 914.08
Point B= 40,20
Max Z= 16x1+16x2
Putting the values;
Z= 960
Max Profit is $ 960
A
Point C= 54,6
Max Z= 16x1+16x2
Putting the values;
Z= 960
Max Profit is $ 960
B
C
O
In this case, we see that the objective function become parallel to first constraint (x1+x2 ≤
60) function, therefore it is the case of multiple optimal solution i.e Point B and C, where the
profit in both the cases is $ 960
Part 3
Suppose Angela and Zooey reconsidered the demand for beef dinners and decided that at
least 20% of their customers would purchase beef dinners.
Here, the objective function will be the same but one of the constraints will change.
Therefore, the new model parameters are:
Max Z= 12x1+16x2
Constraints
x1+x2 ≤ 60………………………(i)
0.25x1+0.5x2 ≤ 20……………(ii)
𝑥1 3
≥
𝑥2 2
2x1-3x2 ≥ 0……………………..(iii)
𝑥2
≥ 0.2
𝑥1 + 𝑥2
-02.x1+0.8x2 ≥ 0………………(iv)
On Solving graphically, we get,
Point O= 0,0
Profit = 0
Point A = 34.28,22.85
Max Z= 12x1+16x2
Putting the values;
Z= 776.96
Max Profit is $ 776.96
Point B= 40,20
Max Z= 12x1+16x2
Putting the values;
Z= 800
Max Profit is $ 800
Point C= 48,12
Max Z= 12x1+16x2
Putting the values;
Z= 768
Max Profit is $ 768
A
B
C
O
In this case, we see that maximum profit is at B, where the number of fish meal is 40 and beef
meal is 20. So, there will be no change in the meal plan of “The Possibility” by changing the
customer demand for beef.