harsh.champ wrote:Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs.2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

(1)240
(2)250
(3)300
(4)360
(5)400

At 200 samosas the revenue = 400
At 220 the revenue = 220*(2 -0.1) = 440 -22 = 418
At 240 = 240 *(2-0.2) = 480 -48 =432
at 300 =300*1.5 = 450
at 360 = 360 *(1+0.2) = 360+72 = 432
at 400 =400*1 =400

The revenue is maximized at 300

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

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

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]

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

_________________________
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??

Sorry my mistake, even I would like to go for hit and trial.......Thanks ajith for pointing it out