To find the maximum volume find the derivative of (2) with respect to r and equate it to zero. 2(1074)ar – 67°r² %3D dr → 2(1074) ar – 6x²r = 0 → ra(2148 – 6ar) = 0 2148 358 →r = 0 or r = 67 Now to find height substitute value of r in h = 1074 – 2ar, 358 h = 1074 – 2n-: = 358. And h=y y = 358. 358 Using r = 98 in 2ar = x we get x-716. Therefore dimensions of a sheet of metal such that the cylinder is created with maximum volume are x=716 mm and y= 358 mm.
To find the maximum volume find the derivative of (2) with respect to r and equate it to zero. 2(1074)ar – 67°r² %3D dr → 2(1074) ar – 6x²r = 0 → ra(2148 – 6ar) = 0 2148 358 →r = 0 or r = 67 Now to find height substitute value of r in h = 1074 – 2ar, 358 h = 1074 – 2n-: = 358. And h=y y = 358. 358 Using r = 98 in 2ar = x we get x-716. Therefore dimensions of a sheet of metal such that the cylinder is created with maximum volume are x=716 mm and y= 358 mm.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 54CR
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Can someone explain to me the math on how r=2148/6pi was found?
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