What is Minimization?
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in terms of global or local range. The definition of minimization in the thesaurus is the process of reducing something to a small amount, value, or position. Minimization (noun) is an instance of belittling or disparagement.
- Approximation - Traditional algorithmic tools cannot solve many real-world algorithmic problems efficiently. When we don't need the exact solution to a problem, but rather approximate value, an approximation algorithm is used. An approximation algorithm is a type of algorithm that returns near-optimal results.
- Local Minimum - A local minimum of a function is a point where the function value is less than that of nearby points but may be greater than that of a distant point.
- Global Minimum - A global minimum is a point where the function value is less than it is at all other possible points.
- Magnification - Magnification is the process of enlarging something. It also applies to increasing the apparent size rather than the physical size. This expansion is undeniably quantifiable.
One major application of calculus is for computing the minimum or maximum value of a function. For example, enterprises frequently want to minimize production costs or maximize profit. In manufacturing, it is regularly necessary to minimize the amount of material used to package a product of a certain size.
The fundamental idea that makes calculus useful in understanding problems of maximizing and minimizing things is that the tangent is horizontal at the peak of a function's graph or the bottom of a trough. That is, at points x0 where f(x0) is a maximum or a minimum, the derivative f′(x0) is 0. Take, for example, the function sin x. This function can have a maximum value of +1 and a minimum value of -1. Its global maxima and minima will be as follows. The global maximum value is 1 and the global minimum value is -1 because sin x has a maximum value of +1 and a minimum value of -1 over all the values for which it is defined (i.e. -infinity to +infinity).
Assume we want to find the maxima and minima of sin x over the range [0,90] (degree). This is a case of local maxima and minima because we have limited the interval over which the function is defined. In this case, sin x will have a minimum at x=0 and a minimum value of 0. It will have a maximum at x=90 and a maximum value of 1. All other sin x values will be between 0 and 1 within the interval [0,90].
Problem Solving Strategy for Minimization
Stage I: Develop the Function.
The first step is to create a function that represents the quantity that you want to optimize. It can be determined by only one variable.
The steps are as follows:
- Sketch the physical situation.
- Take note of any physical constraints imposed by the physical situation.
- Create an equation that connects the quantity you want to optimize to the relevant variables.
- If necessary, rewrite your equation in terms of a single variable using the other information provided.
Stage II: Maximize/Minimize the Function.
You are now faced with a standard max/min problem to solve.
- Calculate the derivative of your equation with respect to your single variable. Then identify the critical points.
- Determine the maximums and minimums as needed.
- Remember to check the endpoints, if any exist.
- Justify your maxima and minima using reasoning from the physical situation, the first derivative test, or the second derivative test.
- Finally, double-check that you answered the question correctly: Reread the problem and ensure that you are providing the requested value(s), such as an x or y value, coordinates, maximum area, shortest time, or whatever was requested.
Parts of an Optimization (Minimization) Problem
A set of decision variables, an objective function, bounds on the decision variables, and the constraints define a minimization problem. This is how the formulation looks.
A vector (one-dimensional array) of decision variables whose values can be changed to find the best solution is denoted as x = (x1, x2,..., xn). A solution is a collection of values for these decision variables.
The objective is a function of the decision variables, f(x). It returns a single number evaluating a solution, which the optimizer attempts to minimize, depending on the formulation.
A range lbi ≤ xi ≤ ubi, i = 1, 2, ..., n on the decision variables, defining which values are permitted are called the bounds of the variables. These bounds define the set of possible solutions, which is referred to as the search space. A lower and/or upper bound can be assigned to each decision variable. If no bounds are specified, the lower and upper bounds are -∞ and +∞ respectively, that is, there are no bounds.
Constraints, such as g1(x→) ≤ b1, are bounds on the functions of the decision variables. They specify which solutions are feasible.
To explain the technicalities of the concept of minimization, we will solve one actual example.
Statement- A manufacturer wants to make a cylindrical can that will hold 1.5 liters of liquid. Determine the dimensions of the can that will minimize the amount of metal used in its construction.
Answer: To solve an optimization minimization problem in calculus, we need to break it down and develop a strategy.
In optimization problems, always start solving your problem by sketching the situation.
The problem demands us to minimize the cost of the metal used in constructing the can. In the given sketch, each piece of metal is separated which are: the can’s circular top, curved side, and circular bottom. Label the can’s height as h and its radius as r. We need to find the values of h and r in terms of V (volume) that will minimize the cost of constructing the can.
After sketching the picture, the next step is to write an equation for the quantity we want to minimize. To solve this step, you will use your everyday knowledge of geometry. In this problem, for example, we want to minimize the cost of constructing the can, which means we want to use very little metal as possible. Here we want to minimize the can’s surface area. So, the equation for that total surface area will be:
ATotal Surface = A Top +A Bottom +A Curved
Now write an equation for the quantity which needs to be minimized (ATotal Surface) in terms of the relevant quantities (r and h).
Our equation for ATotal has two variables, r and h. We must eliminate one of them in order to proceed. The choice of which to keep and which to eliminate is arbitrary; for our solution here, we choose to keep r. Since we’re choosing to work with r, we need to use other detailed information given in the problem to write h in terms of r so we can substitute for ‘h’ as a variable.
Here we are using liters for volume, as we want length measurements for the radius and height, we will also require the volume in terms of length measurement.
Using the fact that 1 litre = 1000 cm3, we can convert 1.5 litres into a centimetre, 1.5 litres = 1500 cm3.
This will, in turn, give a radius and height in terms of centimeters.
Here in the given problem, the constraint is the volume, and we want to minimize the amount of material used.
This means that what we want to minimize is the surface area of the can. We will include both the walls of the can as well as the top and bottom caps of the can.
The equation for the volume (V) and surface area (A) of a cylinder is then,
V= (π r2) (h) = π r2 h
A= (2πr) (h) =2πr h+2 π r2
Following are the equations for the surface area of the caps of the cylinder, and constraints are given for this problem:
Minimize: A =2πrh+2πr2
Constraint: 1500 =πr2h
In this case, it looks like our best option is to solve the constraint for h and plug this into the area function (A).
h=1500 / πr2
A(r) = 2πr (1500/πr2) +2πr2
A(r) =2πr2+3000 / r
Notice that this formula will only make sense from a physical standpoint if r>0, which is a good thing as it is not defined at r=0.
Next, let’s get the first derivative.
A′(r) = 4πr – 3000 / r2
=(4πr3− 3000) / r2
From the above equation, we can find that we have one critical point:
r ≈ 6.2035 (when the derivative is zero)
Note that r=0 is not a critical point because the area function does not exist there, which makes sense from a physical standpoint as well, given that we know that r must be positive in order to actually have a can.
So, we have a single critical point to deal with here and observe that 6.2035 is the only value for which the derivative will be zero and hence the only place with r>0, of course, that the derivative may change sign. It is very easy using test points to check that if 0<r<6.20350, then A′(r)<0 and similarly if r>6.2035 then A′(r)>0. The variant of the first derivative test above then tells us that the absolute minimum value of the area (for r>0) must occur at r=6.2035.
All we need to do is determine the height of the can, and we’ll be done.
h=1500/ π (6.2035)2 ≈ 12.4070
Therefore, if the manufacturer makes the can with a radius of approximately 6.2035 cm and a height of approximately 12.4070 cm, the least amount of material will be used to make the can.
- Most students do not understand that they need to complete two distinct stages as developing the function and minimizing the function.
- Students need to be very good at basic geometry to solve optimization minimization problems. Writing incorrect formulas or equations gives incorrect results.
- Accuracy in math is also significant as a small mistake creates enormous challenges.
The formulas used in the above example are:
Volume of cylinder = πr2h
Total surface area of a cylinder = 2πrh +2πr2
Power rule of derivative:
Here h and r are respectively the radius and height of a cylinder, x is a variable and m is a real number.
Context and Applications
This topic is significant in the professional exams for both undergraduate and graduate courses, especially for
- 10th,11th,12th Standard.
- Engineering Students
- Finance Professionals
- Minimization plays an important role in the entire industrial field.
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