You manufacture two products, A and B, each of which you sell for $1 profit. Product A requires 5 blobs and 3 globs, and product B requires 3 blobs and 5 globs. Your supplier has 120 blobs and 120 globs available.
Use the Excel Solver (or graphical analysis) to answer the following questions.
Starting with the original scenario, where A and B each make $1 profit, your supplier has up to 120 blobs and 120 globs available, but you can only store 200 total blobs and globs:
13. What is the most profit you can make?
14. What solution(s) produce maximum profit? What is the minimum number of each A and B that you must produce to achieve this profit?
15. Would you pay up to $5 for additional storage space for 40 blobs and globs?
16. If the profit on B increases to $2, is your optimal solution constrained by storage space?
17. Can you obtain more blobs and globs for product B by producing negative quantities of product A?
If the company produce X units of A and Y units of B and each one yields a profit of 1
Objective function that maximises the profit is
Max ( X+Y)
such that
5X+3Y<=120 ( constraints related to blobs)
3X+5Y<=120 ( constraints relatd to glob)
Total of blobs / globs stored can be less than or equal to 200, which means
5X+3Y+3X+5Y<=200
X+Y<=25
The graph below shows the equation
The points of optimality are ( 0,24)(24,0) (2.5, 22.5) and (22.5,2.5)
Since both parts yield the same profit, the value of objective function will be the same at all four points ( It is not possible to make a fraction of item, hence the quantity made will be (22,2) or (2,22)
The company should make either 2X and 22Y or 2Y and 22X. The profitability will be same.
(iii) If the storage space increases to 240, the equation will be
8X+8Y<=240
X+Y<=30
Now, the points of intersection will be ( 15,15) and the new graph will be
The new points of optimality will be (0,24)(24,0) and (15.15)
Max profit will be at (15,15) which is
15+15=30
Since the profit increases by 6 with an investment of 5, the offer should be accpeted.
If the profit on B increases by 2 , the new objectie function will be
X+2Y
Value of OF
at (24,0) = 24
at(0,24)=48
at(15,15)= 15+30=45
The optimal solution is not constrained by the space now.
Note: As per policies, I can answer first 4 parts of a question. Inconvenience is regretted.
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