Dimensions of Cylinder that Minimize Cost | Calculus Optimization Problem

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What are optimization problems?

Optimization, by its very name, implies an efficiency of some type: least amount of work, largest volume, smallest distance, greatest profit, etc. If these quantities could be expressed as functions, then graphed, we would be looking for the highest (or lowest) points on the graphs.

Due to the diverse nature of optimization problems, it is difficult to create an algorithm that can be used to solve all problems. They can, generally, be solved by a four step method, outlined as follows:

Step 1: Create a model for the problem.

Define the relationships we are given. Determine what function we are trying to minimize or maximize. This will be the function for which we will have to take the derivative, so we will need to make sure this function has one variable only. Usually this will mean using the given information to make substitutions.

Step 2: Be aware of any constraints on the problem.

We will need to check for boundaries in which our solution may lie. Usually these can be calculated by determining the highest and lowest allowable values for our variable. It is important to note that not all optimization problems have boundaries. In some cases, you may even find it easier to consider these first.

Step 3: Determine the critical points.

Take the derivative of the function and determine the values for which the derivative is 0. Use a sign chart or the second derivative test in order to ascertain whether the critical points are minimum or maximum points.

Step 4: Compare all points.

The function will have a maximum or a minimum at either the endpoints or the critical points. To check, simply compare the values of the function.

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[Note: If you have a closed interval, use the Closed Interval Method. If you have an open interval, you must demonstrate that the point you have found is indeed a maximum or a minimum.]

Question:

Consider a cylindrical tin can which is to be constructed by joining the ends of a rectangular piece of metal to form the cylindrical side, and then attaching circular pieces to form the top and bottom. There are seams around the perimeter of the top and bottom and there is one seam down the side surface (where the ends of the rectangle join together).

Suppose the volume of the can is 1000 cm^3. Also, suppose that the cost of the material is $1.00 per m^2 and the cost of the seam is $0.20 per meter. Find the dimensions of the can that will minimize the cost.

Helpful formulas:

Surface Area of Cylinder=2πrh+2πr^2

Volume of Cylinder=πr^2 h