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Unformatted text preview: (5) A company would like to construct a cylindrical can with an open top with a volume of 4 cubic feet. If the cost of the materials for the sides of the can is $2 per square foot, and the cost of materials for the bottom of the can is $5 per square foot, what are the dimensions of the can (radius, height) which will minimize the cost of the materials? Potentially useful formulas: Volume of a cylinder = r 2 h Surface area around a cylinder = 2 rh Area of a circle = r 2 Find the dimensions of the can (radius, height) which achieve the minimum cost of materials. Solution. In this problem we are trying to minimize the cost of the materials. We can find a formula for this cost as: (cost) = (cost of sides) + (cost of bottom) The cost of the sides is 2(area of sides) and the cost of the bottom of 5(area of bottom). In particular, we have (using the formulas given): C = 5 r 2 + 2 2 rh = 5 r 2 + 4 rh where r is the radius of the can, h is the height, and C is the total cost. Now, we need to eliminate one of these variables in order to get just one variable. To do this, we use the fact that the volume is fixed at 4 cubic feet. Using the formula above this says: 4 = r 2 h We can then solve this for h to get: h = 4 r 2 Substituting this into the equation for C , we obtain C as a function only of r : C = C ( r ) = 5 r 2 + 4 r 4 r 2 = 5 r 2 + 16 /r Now we need to determine the interval on which this function is defined that is to say, we need to find which values of r which satisfy the constraints of the problem. In this case, or course, r certainly cannot be negative. Besides this, the only constraint we actually have is that the volume of the can must be equal to 4. This means that it doesnt make sense to have the radius equal to 0 (since such a can would have no volume). On the other hand, we can make the radius very small, since by then making the can very tall, we would still be able to have the required volume of 4. This says so far: < r Now we need to ask how large r could be. If we made r very large, so that the base is very wide, we ask whether we might still be able to ensure that the volume is 4. We 1 can do this by making the can sufficiently short. Therefore we can make the radius as big as we like. This says that the interval is: (0 , ) Therefore we cannot use the closed interval method, but instead must use the first derivative test (i.e. make a sign chart for the derivative). The next step is to find the critical points for C ( r ). We first take the derivative: C ( r ) = 5 2 r + 16 ( 1) r 2 = 10 r 16 /r 2 The only place this is not defined would be at r = 0, but this is not in the interval, so we dont think of this as a critical point. We therefore consider whereso we dont think of this as a critical point....
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This note was uploaded on 12/10/2011 for the course MATH 2200 taught by Professor Kazez during the Fall '08 term at University of Georgia Athens.
 Fall '08
 KAZEZ
 Calculus

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