The surface given by Volume area of the cylinderical tank is SA = The cost of steel Given 2пяв+2пяч of cylinderical tank V from Ⓒ (3) ), V = 2000 cubic meters. с C = 0·5 (21 ³²²) + 0.3 (21h) пять 2000 = πT 1²h solve for h = ((r) = 2000 1772 h @ substitute the value of cost function ( = 0-5 (217-2²) +0.3 (21Tr (2000) C Tr² 1200 h in , we get 2 + 2
The surface given by Volume area of the cylinderical tank is SA = The cost of steel Given 2пяв+2пяч of cylinderical tank V from Ⓒ (3) ), V = 2000 cubic meters. с C = 0·5 (21 ³²²) + 0.3 (21h) пять 2000 = πT 1²h solve for h = ((r) = 2000 1772 h @ substitute the value of cost function ( = 0-5 (217-2²) +0.3 (21Tr (2000) C Tr² 1200 h in , we get 2 + 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
- -analyze the extreme points and critical points to determine the optimum value that solves your problem.
- -end your problem with a concluding statement
- -find the equation of the tangent at your optimum point
- -create a graph that presents: your function, the derivative and the tangent at the optimum point.
number it so its easier and the prior info is given in the pics, can u please also create the interval table missing in the pics
Transcribed Image Text:Given
An open top cylinderical tank Volume
Cost of the steel used for the top and bottom
meter
= 0.5 dollars
per square
of the steel used for the sides
Cost
2000 cubre meters
0.3 dollars per square meter
Transcribed Image Text:The surface
given by
Volume
The cost of steel
area of the cylinderical tank is
SA = 2πrh+ 2πr²
Given
of cylinderical tonk V
from 3
V = 2000 cubic meters.
)
C = 0·5 (21 ³²²) + 0.3 (21h)
с
пять
2000 = πT 1²h
solve for h
2000
Tr2
h =
c(r) =
substitute the value of
h in , we get
cost function ( = 0.5 (217-2²) +0.3 (2πr (200²))
пяч
-
+
1200
r
(4)
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