[Maths Class Notes] on Optimization Pdf for Exam

Mathematical optimization or optimization means to select the feasible element that depends on a specific standard from a set of available options.

A specific optimization problem includes minimizing or maximizing real functions efficiently by selecting input values within a given set and calculating the function’s value. It is applied in numerous areas of mathematics for specifying the theory of optimization. Optimization means examining “best available” values of the specific objective function in a defined domain including multiple types of objective functions. This article will define what is optimization, Mathematical optimization problems, why use Mathematical optimization etc.

Mathematical Optimization Problems

Now, let us look at some optimization problems. Here, you need to look for the highest or the smallest value that can be considered as a function. The constraint will be normal that can be specified by an equation.

The constraint is the quantity that has to be valid regardless of the solution. You will be looking at one quantity that is clear and has a constant value in every problem. Once you will clearly recognize the quantity to be optimized, it’s not so problematic to calculate it further.

Why use Mathematical  Optimization?

Optimization is a mathematical approach that considers all the factors that influence business decisions. Optimization means careful modeling of the business, a process which itself provides valuable information. The benefits of mathematical optimizations are operational efficiency, cost minimization, performance assessment, and understanding the effects of the variation made in input data.

Important factors included in the optimization are:

1. Decisions–  These are the things that can vary, the things we need to choose upon. For example, the number of products that can be made, how to make the product and dispatch it.

2. Constraints- These designs are the limitation of our decisions. For example, in a logistic problem each mode of transportation has maximum speed and payload, Operations can be controlled for many hours in a day.

Mathematical Optimization Problems in business

Here are some examples of mathematical optimization which will help you to know how mathematical optimization is helpful in business.

1. Portfolio Management– Mathematical optimization helps the business to manage its portfolio. With this, the entrepreneur can decide what stocks and number of stocks should be included in a portfolio that provide maximum return and minimize risk.

2. Stock Level Management –  What stock the business should maintain and when to meet the overall cost of the stock but still meet the required supply SLAs.

3. Hotel Business- What should be the feasible price of the room that will maximize occupancy while considering room availability but staying within a range of prices and considering estimated take-up for the range of possible prices?

Solved Examples

1. Determine two positive numbers whose sum is 300 and whose product is maximum.

Step 1. The first step is to write the equation which will describe the situation.

Let us take two number p and q whose sum is 300 

p + q = 300

Now we will maximize the product

A = pq 

Step 2. Now, we will solve the constraint and substitute this in the above equation

q = 300 – p    A(p) = p(300 – p) =  300p – p²

Step 3. Now we will find the critical points for the equation

A’(p) = 300 – 2p       300 – 2p = 0      p = 150

Step 4. As we got a single value and we can’t assume that this will provide us a maximum product. We will examine to see whether it will give us  maximum value

There are multiple methods to verify this ,but in this case, we can quickly see that

 A’’ (p)= -2

With this, we can conclude that the second derivative is also negative and so A(p) will always concave down and the critical point which we got in step 3 must be relatively maximum and can be a value that gives us a maximum product.

Step 5. And in this step, we will determine the value of y  as we already have the value of x and that can be easily done from the constraint.

q = 300 – 150= 150

Hence, the final answer is  p= 150 and q = 150

2. Let p and q be two positive numbers such that p + 2q and (p+1) (q+2) is a maximum.

Solution:

Step 1. As we have been given constraints of the above problem, it will be represented as

 P + 2q = 50

We are asked to maximize the equation

 

f = (p +1) (q +2)

Now we will solve the constraints for p and q and substitute this into an product equation

p= 50- 2q →   f(q) = (50 – 2q + 1)(q + 2)    =    (51- 2q)(q + 2)    = 102 + 47q – 2q²          

Step 2. In this step, we will find the critical point for the equations.

f’(q) = 47- 4q    →    47-  4q = 0

Step 3. As we got a single value and we can’t just assume that will provide us a maximum product. We will quickly check whether it will give us maximum value.

f’’(q) = – 4

With this, we can conclude that the second derivative is also negative and so f (p) will always be concave down and the critical point which we got in step 2 must be relatively maximum and can be a value that gives us a maximum product.

 

Step 4. Finally, in this step we will answer the question. We are required to provide both the values. As we already have q so we need to find p and that can be easily done from the constraint.

p = 50 – 2(47/4) = 53/2

The final answer to  the question is 

p= 53/2  and  q = 47/4

Quiz time

1. Which of the following is not a step to solve the optimization problem?

1. Write down an equation

2. Answer the problem

3. Find the minimum or maximum value

4. Construct a detailed graph

Fun Facts