ajith wrote:Mom4MBA wrote:It can also be done using Maxima and Minima also;
The formula of revenue will be R = n[2 - 0.10(n-200)/20] , where n is number of samosas
R = n[2 - 0.10(n-200)/20] ....................eq� (i)
simplify it
R = 2n - (n²-200n)/200 ........................eq� (ii)
R' = 2 - (2n-200)/200 , first derivative of R
make R'=0, we will get n=300
R'' = -(2)/200 as this is negative the value of n at R'=0 is the maximum value
put n=300 in eq� (i) we get R=450
so maximum revenue is Rs 450 for a box of 300 samosas
Strictly speaking, it is a step function and thus not continuous and maxima and minima cannot be applied.
In this case you have assumed that 10 paise decrease is gradual where in the actual decrease is discontinuous and happens only in intervals of 20
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Thanks ajith for pointing out the mistake.
Many a times people tend to make functions without even realizing its continuity and its limits.
Mom4MBA wrote:
well isn't it a continuous step function.............
I think not, in your case the rates for 201 samosas and 219 samosas are different but, in actual case, it is same, the change occurs at 220 - hence called step function.
What would you do if your optimization technique gives a maxima at 291? [in this case it works perfectly and gives a maxima at a step which is 300]
Also,we can see over here that by formulating the f:n and calculating the maxima and minima,the soln. process becomes quite cumbersome.
I would like to go by the hit-and-trial method in this case. What-say guys??
