Can someone explain to me the math on how r=2148/6pi was found? Step 3To find the maximum volume find the derivative of (2) with respect to r and equate it to zero.dv= 2(1074)xrtr – 6x²r²drdv||dr= 2(1074) Tr – 6x²r² = 0= rT (2148 – 6r) = 02148358» r = 0 or r =блNow to find height substitute value of r in h = 1074 – 2ar,358h1074 – 2n-358.||And h=y = y = 358.358Using r =in 2ar = x we get x=716.Therefore dimensions of a sheet of metal such that the cylinder is created with maximum volume arex=716 mm and y= 358 mm.

Given: rπ(2148-6πr)=0To explain how the value of r=21486π came from the above equation

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: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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