Science:Math Exam Resources/Courses/MATH110/April 2018/Question 10
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Question 10 

You decide to manufacture a can that contains a fixed volume of by bending a rectangular sheet of metal to form a cylinder and then soldering a circular disk at each end so that your can has both a top and a bottom. You use a metal that cost $1 per for the side and a different metal that costs $2 per for the top and bottom. Find the radius of the can which will minimize its manufacturing cost. Show that your answer corresponds indeed to an absolute minimum. Recall the volume of a cylinder of height and base radius is and its surface area is . 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Write the total cost as a function of radius and height. Then use the volume constraint to eliminate one of the variables. 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. We can split the surface area into , where is the surface area of the side of the can, and is the area of the top and bottom of the can. Then we have and . Now we can calculate the total cost of manufacturing a can by calculating the cost of the sides of the can and the cost of the top and bottom of the can separately. If is the total cost in dollars of manufacturing one can, then we have since the metal used for the side costs $1 per and the metal for the top and bottom costs $2 per .
Next, we will make use of the given volume constraint, which can be stated as . Solving for (and noting that we can divide through by since must be positive), we get This will allow us to rewrite the cost function as a function in only one variable.
Substituting into the cost formula gives To minimize this function, we will find the roots of its first derivative.
By the power rule, we get Setting this to zero and rearranging the equation gives Multiplying through by and diving by leaves It follows that Note that the first derivative of the cost function had only one root, so to show that it is an absolute minimum it suffices to show that it is a local minimum (since the endpoints of the domain are not in consideration). Moreover, since the first derivative has no other roots, we can show that the point is a local minimum by simply checking the value of the first derivative of the function at some other point in the domain. We will check the value of the first derivative of at the point (which is smaller than ). At , the first derivative of is This shows that to the left of the local extremum the function is decreasing, so the radius we found is indeed a local minimum and hence the absolute minimum.
Answer: The radius that minimizes the manufacturing cost is . 