It was shown in class that the closed cylindrical can of maximal volume that may be constructed to have surface area: SA units² must have radius: r = √6π =(2x) Thus, the dimensions of the cylindrical can will be optimized if and only if = 2. = TU Now suppose that the following additional constraint is imposed on this optimization problem: The curved side and the circular top and bottom of the cylindrical can must be cut from rectangular sheets of metal. Thus, to minimize waste in the construction process, the side will be formed from a single rectangular sheet having dimensions: 2 лrxh and the discs for the top and bottom will be cut from another sheet of metal having dimensions: 2 rx 4r. Assume that the excess material from the second sheet used to form the circular bases must be scrapped and cannot be reused. h = ² xr In this scenario, use calculus to prove that the cylinder of largest volume that can be constructed out of the two rectangular sheets having a combined surface area of: SA = 8 r² + 2лrh must have base radius r and height h related by the formula: h r h units and height: h= (2x ≈ 2.55 2 ar units. SIDE h = 2r TOP BOTTOM
It was shown in class that the closed cylindrical can of maximal volume that may be constructed to have surface area: SA units² must have radius: r = √6π =(2x) Thus, the dimensions of the cylindrical can will be optimized if and only if = 2. = TU Now suppose that the following additional constraint is imposed on this optimization problem: The curved side and the circular top and bottom of the cylindrical can must be cut from rectangular sheets of metal. Thus, to minimize waste in the construction process, the side will be formed from a single rectangular sheet having dimensions: 2 лrxh and the discs for the top and bottom will be cut from another sheet of metal having dimensions: 2 rx 4r. Assume that the excess material from the second sheet used to form the circular bases must be scrapped and cannot be reused. h = ² xr In this scenario, use calculus to prove that the cylinder of largest volume that can be constructed out of the two rectangular sheets having a combined surface area of: SA = 8 r² + 2лrh must have base radius r and height h related by the formula: h r h units and height: h= (2x ≈ 2.55 2 ar units. SIDE h = 2r TOP BOTTOM
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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Transcribed Image Text:It was shown in class that the closed cylindrical can of maximal
volume that may be constructed to have surface area: SA units²
must have radius: r =
√6π
=(2x)
Thus, the dimensions of the cylindrical can will be optimized
if and only if = 2.
=
TU
Now suppose that the following additional constraint is imposed on this optimization problem:
The curved side and the circular top and bottom of the cylindrical can must be cut from
rectangular sheets of metal. Thus, to minimize waste in the construction
process, the side will be formed from a single rectangular sheet having
dimensions: 2 лrxh and the discs for the top and bottom will be cut
from another sheet of metal having dimensions: 2 rx 4r.
Assume that the excess material from the second sheet used to form the
circular bases must be scrapped and cannot be reused.
h = ² xr
In this scenario, use calculus to prove that the cylinder of largest
volume that can be constructed out of the two rectangular sheets
having a combined surface area of: SA = 8 r² + 2лrh
must have base radius r and height h related by the formula:
h
r
h
units and height: h= (2x
≈ 2.55
2 ar
units.
SIDE
h = 2r
TOP
BOTTOM
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