Key Concepts of Optimization
Optimization is the process by which a variable is maximized or minimized in relation to another for real-life purposes. It allows one to optimize a target measure, such as profit or volume, using its constituent parts, such as production or length, while working with constraints, such as costs of production or available material. Often, it helps to draw a figure too.
First, one must be given the two equations by which variables are introduced in relation to each other. Let's take, for example, the maximization of the area enclosed within a fence using 800 feet of fencing material.
A=lw
800=2l+2w
The first equation solves for area, while the second gives the relationship between length and width using the perimeter function. The next step is to solve for either independent variable (the variables that are not the goal of maximization or minimization) in terms of the other. In this example, we'll solve for w.
800=2l+2w
2w=800-2l
w=400-l
Now, we can substitute 400-l in for w in the area equation.
A=lw
A=l(400-l)
A=400l-l^2
From here, we take the derivative of this function.
A=400l-l^2
A'=400-2l
Maximums and minimums occur at points where the derivative of the function equals zero. However, just because a point causes the derivative to equal zero, it does not necessarily mean that a maximum or a minimum is present. Let's set A'=0 and solve for l.
0=400-2l
2l=400
l=200
In this case, we only get one l value. We must determine whether this is indeed a maximum. There are multiple ways to determine this, but the fastest is to take the second derivative of the equation, plug in the value(s) obtained from setting the first derivative equal to zero, and determine whether the second derivative at that value is positive (meaning that it is a minimum), negative (meaning that it is a maximum), or zero (meaning that it is neither).
A'=400-2l
A''=-2
In this case, we didn't have to plug 200 in for l, since the second derivative lacks an l value. This means that the second derivative is negative everywhere. So the length that produces the maximum area is 200 ft. Plugging this back into the equation we got solving for w shows that the width is also 200 ft. Putting these together in the area equation shows that the maximum area that can be enclosed within a rectangular fence with 800 ft of fencing is 40,000 sq ft.
Though this situation is handled simply, these methods apply universally. Identify target, identify constraints, reduce target to terms of one variable, find maximum/minimum, show result.
First, one must be given the two equations by which variables are introduced in relation to each other. Let's take, for example, the maximization of the area enclosed within a fence using 800 feet of fencing material.
A=lw
800=2l+2w
The first equation solves for area, while the second gives the relationship between length and width using the perimeter function. The next step is to solve for either independent variable (the variables that are not the goal of maximization or minimization) in terms of the other. In this example, we'll solve for w.
800=2l+2w
2w=800-2l
w=400-l
Now, we can substitute 400-l in for w in the area equation.
A=lw
A=l(400-l)
A=400l-l^2
From here, we take the derivative of this function.
A=400l-l^2
A'=400-2l
Maximums and minimums occur at points where the derivative of the function equals zero. However, just because a point causes the derivative to equal zero, it does not necessarily mean that a maximum or a minimum is present. Let's set A'=0 and solve for l.
0=400-2l
2l=400
l=200
In this case, we only get one l value. We must determine whether this is indeed a maximum. There are multiple ways to determine this, but the fastest is to take the second derivative of the equation, plug in the value(s) obtained from setting the first derivative equal to zero, and determine whether the second derivative at that value is positive (meaning that it is a minimum), negative (meaning that it is a maximum), or zero (meaning that it is neither).
A'=400-2l
A''=-2
In this case, we didn't have to plug 200 in for l, since the second derivative lacks an l value. This means that the second derivative is negative everywhere. So the length that produces the maximum area is 200 ft. Plugging this back into the equation we got solving for w shows that the width is also 200 ft. Putting these together in the area equation shows that the maximum area that can be enclosed within a rectangular fence with 800 ft of fencing is 40,000 sq ft.
Though this situation is handled simply, these methods apply universally. Identify target, identify constraints, reduce target to terms of one variable, find maximum/minimum, show result.
