Cylinder Volume and Surface AreaDate: 04/14/99 at 21:18:13 From: Long Nguyen Subject: Maxima and Minima - Cylinder Volume Hi, My name is Long and I stumbled across a question that I just can't figure out. Can you please help? You have to make a cylinder that holds the greatest volume when the entire surface area of the cylinder (including the area of both the circles on the top and bottom) is 600 cm squared. You can make the cylinder any way you want but it has to be a cylinder. I am having a lot of difficulty with this and I hope you can help. Please explan to me how to solve this and please show me in mathematical terms, and can you tell me how we would know that the answer you give me is the maximum volume it can hold? Thank you very much, Long Nguyen Date: 04/16/99 at 16:00:17 From: Doctor Jeremiah Subject: Re: Maxima and Minima - Cylinder Volume Dear Long: It's a hard question. Figuring out the maximum volume requires calculus. The surface area of a cylinder is A = 2*Top + Side A = 2*CircleArea + Rectangle A = 2*CircleArea + CircleCircumference*Height 600 = 2*(pi*r*r) + (2*pi*r*h) The Volume of a cylinder is V = Top*Height V = CircleArea*Height V = (pi*r*r)*h To find the maximum volume we must have one variable (r), so we must solve the surface area for h and substitute. A = 2*Top + Side A = 2*CircleArea + Rectangle A = 2*CircleArea + CircleCircumference*Height 600 = 2*(pi*r*r) + (2*pi*r*h) 600 - 2*(pi*r*r) = 2*pi*r*h (600 - 2*pi*r*r)/(2*pi*r) = h V = (pi*r*r)*h <== h = (600 - 2*pi*r*r)/(2*pi*r) V = (pi*r*r)*(600 - 2*pi*r*r)/(2*pi*r) V = (pi*r*r)*600/(2*pi*r) - (pi*r*r)*(2*pi*r*r)/(2*pi*r) V = (2*pi*r)*r*300/(2*pi*r) - (2*pi*r)*r*(pi*r*r)/(2*pi*r) V = r*300 - r*(pi*r*r) V = 300r - pi*r^3 Now to find a minimum or maximum of V we must set V to 0. A zero slope means a maximum or a minumum. Then differentiate with respect to r. dV/dr = d(300r - pi*r^3)/dr After differentiating you know what r equals. Plug that into the Volume equation V = 300r - pi*r^3 and find the maximum volume. If you need more help, please write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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