"Optimization is a mathematical discipline that concerns the finding of minima and maxima of functions" How To Solve the Optimization Equation The constraint and optimizing equations are now known; 2x+3y=1000 m. and xy=A. Notice that this is a system of equations. To solve this system, one will need to get the equation xy=A in terms of all x or all y. In this example, the y variable is used (though it would be perfectly fine to use the x instead). Therefore, the y variable in the equation 2x+3y=1000 m. will be isolated as follows: 2x+3y=1000 2x=1000-3y x=(1000-3y)/2 x=500-3y/2 The equation x=500-3y/2 can now be substituted back into the optimizing equation as follows:
xy=A
(500-3y/2)y=A
500y-(3y^2)/2=A
Work Cited: www.friartuck.net Example problem cited: www.suite101.com
In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval. In this section we are going to look at another type of optimization problem. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be easily described by an equation, but in these problems it will be easy to deal with as we’ll see. The first step in all of these problems should be to very carefully read the problem. Once you’ve done that the next step is to identify the quantity to be optimized and the constraint. In identifying the constraint remember that the constraint is something that must true regardless of the solution. In almost every one of the problems we’ll be looking at here one quantity will be clearly indicated as having a fixed value and so must be the constraint. Once you’ve got that identified the quantity to be optimized should be fairly simple to get. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first! Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area. Solution In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that. In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are, Maximize: A = xy Constraint: 500 = x + 2y
Optimization allows you to find the maximum or minimum of any problem. This process requires the use of calculus and may be difficult at times. The optimizing process allows you to solve problems such as the volume of three dimensional objects and the amount of certain materials to use in order to lower production costs. You can solve these problems in two dimensions.
Set your variables accordingly and write an equation. Make sure to minimize your variables to just two that relate to one another.
Example: 2y = 8x^2+12x
Isolate one of the variables (usually the one that goes on the Y axis):
In mathematics, computer science and economics, optimization, or mathematical programming, refers to choosing the best element from some set of available alternatives.
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by systematically choosing the values of real or integer variables from within an allowed set. This formulation, using a scalar, real-valued objective function, is probably the simplest example; the generalization of optimization theory and techniques to other formulations comprises a large area of applied mathematics. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.
Solution
In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that.
In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,
"Optimization is a mathematical discipline that concerns the finding of minima and maxima of functions"
ReplyDeleteHow To Solve the Optimization Equation
The constraint and optimizing equations are now known; 2x+3y=1000 m. and xy=A. Notice that this is a system of equations. To solve this system, one will need to get the equation xy=A in terms of all x or all y. In this example, the y variable is used (though it would be perfectly fine to use the x instead). Therefore, the y variable in the equation 2x+3y=1000 m. will be isolated as follows:
2x+3y=1000
2x=1000-3y
x=(1000-3y)/2
x=500-3y/2
The equation x=500-3y/2 can now be substituted back into the optimizing equation as follows:
xy=A
(500-3y/2)y=A
500y-(3y^2)/2=A
Work Cited: www.friartuck.net
Example problem cited: www.suite101.com
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ReplyDeleteIn optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval.
ReplyDeleteIn this section we are going to look at another type of optimization problem. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be easily described by an equation, but in these problems it will be easy to deal with as we’ll see.
The first step in all of these problems should be to very carefully read the problem. Once you’ve done that the next step is to identify the quantity to be optimized and the constraint.
In identifying the constraint remember that the constraint is something that must true regardless of the solution. In almost every one of the problems we’ll be looking at here one quantity will be clearly indicated as having a fixed value and so must be the constraint. Once you’ve got that identified the quantity to be optimized should be fairly simple to get. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first!
Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.
Solution
In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that.
In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,
Maximize: A = xy
Constraint: 500 = x + 2y
workcited:http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
Optimization allows you to find the maximum or minimum of any problem. This process requires the use of calculus and may be difficult at times. The optimizing process allows you to solve problems such as the volume of three dimensional objects and the amount of certain materials to use in order to lower production costs. You can solve these problems in two dimensions.
ReplyDeleteSet your variables accordingly and write an equation. Make sure to minimize your variables to just two that relate to one another.
Example:
2y = 8x^2+12x
Isolate one of the variables (usually the one that goes on the Y axis):
y = 4x^2+6x
Set the derivative of the equation to 0:
f'(y) = 8x+6
0 = 8x+6
Solve for the remaining variable:
x = -(3/4)
citation: http://www.ehow.com/how_5948647_solve-calculus-optimization-problems.html
In mathematics, computer science and economics, optimization, or mathematical programming, refers to choosing the best element from some set of available alternatives.
ReplyDeleteIn the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by systematically choosing the values of real or integer variables from within an allowed set. This formulation, using a scalar, real-valued objective function, is probably the simplest example; the generalization of optimization theory and techniques to other formulations comprises a large area of applied mathematics. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.
Solution
In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that.
In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,