Question

3. This problem revisits the question from Problem Set I about finding the dimensions of a one-liter can that will have the minimum cost. We will now approach it using calculus methods. In class, we found the dimensions of a right circular cylinder (a “can") that has a volume of 1,000 cm3 using the minimum possible material. This assignment changes that problem slightly by seeking the minimum cost for a right circular cylinder whose volume is 1,000 cm3 where the costof materials for the bottom, top, and side are different Suppose the materials for the bottom of your cylinder cost 0.6 cents per square centimeter the materials for the top cost 0.5 cents per square centimeter, and the materials for the side cost 0.3 cents for square centimeter A. (One point.) Write a function for the total cost of the cylinder in terms of its radius (r) and ts height (h) B. (Two points.) Write an equation expressing the 1,000 cm3 volume in terms of the radius and height. Solve your equation for either r or h and subsfitute the result into your cost function C. (Three points.) Differentiate your cost function from Part B and determine where its minimum value occurs D. (One point.) The x-coordinate of the minimum is the value for r or h (whichever you ended up with in your cost function in Part B) that produces the cylinder of minimum cost. Now use your work in Part B to find the other, and state the radius and height of the cylinder of minimum cot Round both values to the nearest tenth of a centimeter

Answer 1

lvder : Cost= 0,5 onts pe Slde (a) Base Area .- 41 2 (Tr*) Co. 6) Cost for base ons = (bi Volume T looo LooO - TT6

Cc) To Rnd winimun, we fund dy and equati 2 do =๐づ a.afiv = 60C dr а-атт 22T* 4-43 cm Raduis that produ<as miunimum Cosi 4.4 ht Minimum Cost .. 1-11(4.49t Go으 4.43