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Keep in mind that we do not and will not have any sort of vetting procedure for responders. Accept answers at your own risk. We use automoderator for a few things. It may delete a post erroneously. We ban all bots. I have done many optimization problems but I am completely stuck on how to set up this particular problem. My teacher needs me to solve a word problem depicting a plastic cup in which I have the volume, but no other information.
The exact problem is as follows:. Pizza Hut wants to produce a right circular cylindrical cup open top that holds cubic centimeters of liquid when filled to the top. The Pizza Hut Company wants you to design a cup that can be constructed with the least amount of material, so they can cut back on production costs. What are the dimensions of the cup? I know you need a couple of different equations, and maybe you need to utilize the pythagorean theorem?
If somebody could just explain how to set the problem up, I can easily figure the rest out. Also, even if we can find the endpoints we will see that sometimes dealing with the endpoints may not be easy either. 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. 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. However, suppose that we knew a little bit more information.
Nowhere in the above discussion did the continuity requirement apparently come into play. Also, the function is always decreasing to the right and is always increasing to the left. There are actually two ways to use the second derivative to help us identify the optimal value of a function and both use the Second Derivative Test to one extent or another. 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.
Suppose that we are looking for the absolute maximum of a function and after finding the critical points we find that we have multiple critical points. We could do a similar check if we were looking for the absolute minimum. Doing this may not seem like all that great of a thing to do, but it can, on occasion, lead to a nice reduction in the amount of work that we need to do in later steps.
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. In fact, we will have the same requirements for this method as we did in that method.
As we work examples over the next two sections we will use each of these methods as needed in the examples.
In some cases, the method we use will be the only method we could use, in others it will be the easiest method to use and in others it will simply be the method we chose to use for that example. With some examples one method will be easiest to use or may be the only method that can be used, however, each of the methods described above will be used at least a couple of times through out all of the examples.
We may need to modify one of them or use a combination of them to fully work the problem. There is an example in the next section where none of the methods above work easily, although we do also present an alternative solution method in which we can use at least one of the methods discussed above.
Next, the vast majority of the examples worked over the course of the next section will only have a single critical point. Problems with more than one critical point are often difficult to know which critical point s give the optimal value. There are a couple of examples in the next two sections with more than one critical point including one in the next section mentioned above in which none of the methods discussed above easily work.
In that example you can see some of the ideas you might need to do in order to find the optimal value. This was done to make the discussion a little easier. We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft 3.
Note as well that the cost for each side is just the area of that side times the appropriate cost. As with the first example, we will solve the constraint for one of the variables and plug this into the cost. Now we need the critical point s for the cost function. We are constructing a box and it would make no sense to have a zero width of the box. Secondly, there is no theoretical upper limit to the width that will give a box with volume of 50 ft 3.
The third method however, will work quickly and simply here. Also, even though it was not asked for, the minimum cost is: This example is in many ways the exact opposite of the previous example. In this case we want to optimize the volume and the constraint this time is the amount of material used.
If you can do one you can do the other as well. Note as well that the amount of material used is really just the surface area of the box. In this case we can exclude the negative critical point since we are dealing with a length of a box and we know that these must be positive. Do not however get into the habit of just excluding any negative critical point.
There are problems where negative critical points are perfectly valid possible solutions. Now, as noted above we got a single critical point, 1. In both examples we have essentially the same two equations: 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.
In Example 3, on the other hand, we were trying to optimize the volume and the surface area was the constraint. This is one of the more common mistakes that students make with these kinds of problems. They see one problem and then try to make every other problem that seems to be the same conform to that one solution even if the problem needs to be worked differently.
Keep an open mind with these problems and make sure that you understand what is being optimized and what the constraint is before you jump into the solution. Also, as seen in the last example we used two different methods of verifying that we did get the optimal value. Do not get too locked into one method of doing this verification that you forget about the other methods.
This will in turn give a radius and height in terms of centimeters. In this problem the constraint is the volume and we want to minimize the amount of material used. Here is a quick sketch to get us started off.
Getting Help; Doing Homework; Problem Solving; Studying For an Exam; Taking an Exam; Learn From Your Errors; We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function This section is generally one of the more difficult for students taking a Calculus.
Optimization mean the achievement of result with certain function to the best possible extent. We offer optimization homework help in math.
Optimization Problems in Calculus: Examples & Explanation. Learning Calculus: Basics & Homework Help Calculus: Help and Review Optimization Problems in Calculus. Oct 26, · At which points on the curve y = 1 + 60x3 − 2x5 does the tangent line have the largest slope? (x, y) (smaller x-value)= (x, y) (larger x-value)=Status: Resolved.
Step-by-step solutions to all your Calculus homework questions - Slader. Apr 24, · Forums > Homework Help > Calculus and Beyond Homework > Homework Help: Optimization problem solved two ways (algebra or calculus) Apr 24, #1. VinnyCee. 1. The problem statement, all variables and given/known data So, this is essentially an optimization problem. The wikipedia entry is advanced, but reading the first paragraph should.