Oil Can DimensionsDate: 12/11/2001 at 23:34:29 From: Shannon Subject: Difficult word problem An oil can is in the form of a cylinder with a base 5cm in radius. If the oil can holds one liter, how tall is it? I have already figured out this part of the question; it is the next part that is giving me all the trouble. What are the dimensions of an oil can with a one-liter capacity that uses the least amount of tin? The prof. swears that no calculus is needed to complete the problem, but I beg to differ. Thank you. Shannon Date: 12/12/2001 at 00:31:38 From: Doctor Jeremiah Subject: Re: Difficult word problem Hi Shannon, Here is the method for getting minimums and maximums: The most important thing in order to find minumum area is to get an equation for the area that has only one independant variable. The minimum surface area will be a can with an unknown radius and height. So the values that you have will not work for the minimum and we will have to find new ones. (The ones you have are for a can that does not have the minimum area.) First you find an equation for the surface area. Since it's a right circular cylinder the surface area is: A = 2 * area of base + area of sides A = 2 * area of circle + circumferecece of circle * height A = 2 * Pi*radius^2 + 2*Pi*radius*height But that is not all you need. There are too many variables. We need to get rid of the height. The other clue is that the can always holds one litre, which means we need an equation for volume: V = area of base * height 1000 cm^3 = 1 L = area of circle * height 1000 = Pi*radius^2 * height Now we can use the equation for volume to determine the height and substitute that into the equation for surface area. 1000 = Pi*radius^2 * height height = 1000 / Pi*radius^2 Okay, we have a value for the height in terms of radius. And if we substitute that into the area: A = 2 * Pi*radius^2 + 2*Pi*radius*(1000 / Pi*radius^2) A = 2 * Pi*radius^2 + 2000/radius Now we have an equation for area that will work, because it has only one independant variable (the radius). Think about what the graph of the area will look like. The minimum will be at the bottom of a dip where the slope changes from negative to positive. The absolute bottom of the dip will have a slope of zero. So what we need to do is find out where the slope of this equation is equal to zero. You could actually draw the graph and figure it out. Thats why the professor said you didn't need calculus. Just graph this equation and look for the minimum: A = 2 * Pi*radius^2 + 2000/radius But since I am not going to draw a graph, I *will* use calculus to calculate it this time. Taking the derivative gives the slope, so all we have to do is differentiate the equation for area to get the slope of the area: A = 2 * Pi*radius^2 + 2000/radius If we differentiate it: d/dradius( A ) = d/dradius( 2 * Pi*radius^2 + 2000/radius ) After you take the derivative you end up with: dA/dradius = 4*Pi*radius - 2000/radius^2 Now remember we said that the minimum was when the slope was 0? slope = 0 = 4*Pi*radius - 2000/radius^2 And if you solve that for minimum radius: 5.41926 cm = radius Just to show you it is the right answer, here are some calculations of points near the minimum: radius = 5.40 cm A = 553.588 cm^2 radius = 5.41 cm A = 553.583 cm^2 radius = 5.41926 cm A = 553.581 cm^2 <== minimum radius = 5.43 cm A = 553.583 cm^2 radius = 5.44 cm A = 553.589 cm^2 Let me know if you would like more details on the calculus part. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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