- 17.01.2019

Optimal values are often either the maximum or the minimum values of a certain function. Optimization Problems in Calculus: Steps. Sample problem: Find the maximum area of a rectangle whose perimeter is meters. Note: This is a typical optimization problem in AP calculus. Step 1: Determine the function that you need to optimize. This formula may involve more than one variable. Write any equations relating the independent variables in the formula from step 3. Use these equations to write the quantity to be maximized or minimized as a function of one variable.

Identify the domain of consideration for the function in step 4 based on the physical problem to be solved. Time to use Calculus! That is, the graph of A r versus r is always concave up.

For example, the problem could have asked to find the value of the smallest possible surface area A, or the minimum cost. Stage I: Develop the function. Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps: Draw a picture of the physical situation. 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 rectangular field with a fence. Determine the dimensions of the field that will enclose the largest area. Show 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.

In this problem we want to maximize the area of a field and we know that will use ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. However, if we solve the constraint for one of the two variables we can substitute this into the area and we will then have a function of a single variable. Having these limits will also mean that we can use the process we discussed in the Finding Absolute Extrema section earlier in the chapter to find the maximum value of the area.

That means our only option will be the critical points. So according to the method from Absolute Extrema section this must be the largest possible area, since the area at either endpoint is zero.

In the previous problem we used the method from the Finding Absolute Extrema section to find the maximum value of the function we wanted to optimize. Also, even if we can find the endpoints we will see that sometimes dealing with the endpoints may not be easy either.

Method 1 : Use the method used in Finding Absolute Extrema. This is the method used in the first example above. If these conditions are met then we know that the optimal value, either the maximum or minimum depending on the problem, will occur at either the endpoints of the range or at a critical point that is inside the range of possible solutions.

There are two main issues that will often prevent this method from being used however. First, not every problem will actually have a range of possible solutions that have finite endpoints at both ends. Method 2 : Use a variant of the First Derivative Test.

However, in this case, unlike the previous method the endpoints do not need to be finite. This will not prevent this method from being used.

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Identify the domain of consideration for the function in step 4 based on the physical problem to be solved. It can depend on only one variable. Please comment below!

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This was done to make the discussion a little easier. Assuming that all the material is used in the construction process determine the maximum volume that the box can have. Your first 30 minutes with a Chegg tutor is free! What it does do is allow us to potentially exclude values and knowing this can simplify our work somewhat and so is not a bad thing to do. So it looks like the only critical point will come from determining where the numerator is zero.

Please read and accept our website Terms and Privacy Policy to post a comment. Do not get too locked into one method of doing this verification that you forget about the other methods. One of the main reasons for this is that a subtle change of wording can completely change the problem. In fact, we will have the same requirements for this method as we did in that method.
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**Masar**

This section is generally one of the more difficult for students taking a Calculus course. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test. What size square should be cut out of each corner to get a box with the maximum volume? The second way of using the second derivative to identify the optimal value of a function is in fact very similar to the second method above. However, in Example 2 the volume was the constraint and the cost which is directly related to the surface area was the function we were trying to optimize. Want to see how we solve other example problems?

**Goltigal**

We are constructing a box and it would make no sense to have a zero width of the box. Nowhere in the above discussion did the continuity requirement apparently come into play.

**Mazum**

What it does do is allow us to potentially exclude values and knowing this can simplify our work somewhat and so is not a bad thing to do.