calculus project
Liu 02
Methods of Optimising the Profit of My Father’s Company
· Chapter 1 Background
My father has run his own company since I was born. As time goes by, his company becomes larger and larger and involved in more areas. Thus, managing all his departments well and maximising the total profit is truly essential for his company.
His company had two departments for selling goods initially while presently he manages eight departments merely for sales.
· Chapter 2 Constructing Model
In the first year, only 60 customers were convinced to invest our company and purchased our products. Suppose this two departments share the same profit function: p(x)= (0≤x≤1). For the x represents the number of customers in hundreds and indicates the profit of each product for each purchaser. Unit is in 100 thousand RMB.
Then the total profit can be demonstrated by the equation
P=a* +b* (0≤a, b≤1), where the independent varieties a and b represent the number of customers for department A and department B respectively.
Next, we need to find out the maximum of P. As there are 60 customers in total, then the sum of a+b is 0.6. So that we can use 0.6-a to represent b to make the equation become a one variety function. Therefore
P=a*+(0.6-a)* (0≤a≤0.6)
Following that, we calculate some certain value to test and estimate the greatest value of P.
|
a |
0 |
0.1 |
0.2 |
0.3 |
0.4 |
0.5 |
0.6 |
|
P |
0.5171 |
0.5540 |
0.5747 |
0.5814 |
0.5747 |
0.5540 |
0.5171 |
|
a |
0.29 |
0.3 |
0.31 |
|
P |
0.5813 |
0.5814 |
0.5813 |
From these two charts, I estimate that when a=0.3, b=0.3, the profit reaches the maximum of 58140 RMB. In another way to think about this question, I can know that the data are symmetric by (0.3,0.5814) by observation. Theoretically, the a* and b* are symmetric due to a+b=0.6. Also, I can also prove it in mathematical ways:
I take the first of P= a*+(0.6-a)* (0≤a≤0.6), that is
P’=*+
*
Let P’=0, then I solve that a=0.3.
|
Interval |
0≤a<0.3 |
0.3<a≤0.6 |
|
P’ |
+ |
- |
|
P |
increase |
decrease |
So that when x equals to 0.3, P reaches the maximum.
In all, in order to optimise the profit of our company, we should balance the number of customers for each department as 30 for each. In this way, we can gain the greatest profit as 51840 RMB. What’s more that’s the way my father did.
Then the question comes, what if the number of independent varieties becomes 8? I think I have to find a formula to solve this question in an easy way.
Here comes the Jensen’s . One of its forms I have to apply later is: ≤n*f( when f(x) concaves down in the certain domain and only when x1=x2=…=xn so that these two equations can be equal.
First, P’(x)=*.
Second, p’’(x)=.
As x∈[0,1], (1-)∈[0,1], which means both x and (1-) are positive. Then it is obvious that p’’(x)<0 when x∈[0,1]. Furthermore, p(x) concaves down through the domain of [0,1].
Therefore, we can use this formula, Jensen’s Inequality, to solve the problem above:
p(a)+p(b)≤2*p[(a+b)/2], and 2*p[(a+b)/2]=2*p(0.3)=0.5814(¥100000).
That is to say, p(a)+p(b)≤0.5814 and only when a=b=0.3 so that p(a)+p(b) can reach the maximum of 0.5814.
Finally, I can draw the same conclusion as before: in order to optimise my profit of our company, we should balance the number of customers for each department as 30 for each. In this way, we can gain the greatest profit as 518400 RMB.
· Chapter 3 Application
In 2017 with 8 departments, there were about 800 customers who were convinced to buy products, and the profit of each department can also be modelled by p(x)= (0≤x≤1). For the x represents the number of purchasers in thousands and indicates the profit of each product for each purchaser. Unit is in million RMB.
Then in order to find out the maximum profit of our company, we can utilize the Jensen’s Inequality.
P=≤8*p()=8*p(0.8/8)=0.7976(¥1000000).
So that P≤7976000(¥), only when x1=x2=…=x8=0.1 that P can reach the greatest value of 7976000.
In conclusion, in order to optimise my profit of our company, we should balance the number of customers for each department as 100 for each. In this way, we can gain the greatest profit as 7976000 RMB.
· Chapter 4 Reflection
In this essay, I use the first derivative to solve the maximum and apply the second derivative into the famous Jensen’s Inequality. In the future we can use this formula to solve the extremum for a series of equations shared in a same function and different independent varieties.
Bibliography
1. Peter Crippin, Calculus 12. Nelson Publishing
2. https://www.desmos.com/calculator/rdxofkozwi
3. https://en.wikipedia.org/wiki/Jensen%27s_inequality
http://www.cs.yorku.ca/~kosta/CompVis_Notes/jensen.pdf