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anonymous
 one year ago
please help
calculus: application of maxima and minima
Find the volume of the largest right circular cylinder that can be cut from a circular cone of radius 6 cm and height 9 cm
the V should be 48pi cm^3
anonymous
 one year ago
please help calculus: application of maxima and minima Find the volume of the largest right circular cylinder that can be cut from a circular cone of radius 6 cm and height 9 cm the V should be 48pi cm^3

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what have you tried so far ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks like an optimization problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i've tried to solve it but i don't know what to do and also how to insert the derivatives

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they have the same SA right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0How so ? Maybe, as a start draw a nice sketch of cylinder inside a cone and label the given dimensions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I think, this problem is equivalent to finding the volume of the largest cylinder that can be inscribed inside a cone..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohw now i get the figure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443926956736:dw like this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443926942036:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah there you go lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0That looks good! label the given dimensions also..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0show radius 6 and height 9 in the diagram

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443937323135:dw volume of cylinder \[V = \pi r^2 h\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443927181720:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[V=\pi \left( 6\frac{ 2h }{ 3 } \right)^2 h\] diff. and find stationary values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then find second derivative and see where it is maxima

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1it is a constrained min/max problem. plug stationary points and boundary points into volume formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should simplify it? i'm confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know what to do with the volume part :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dV }{ dh }=\pi[h*2(6\frac{ 2h }{ 3 })*\frac{ 2 }{ 3 }+1\left( 6\frac{ 2h }{ 3 } \right)^2]\] simplify and put dV/dh =0 and find values of h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[=\pi \left( 6\frac{ 2h }{ 3 } \right)\left[ \frac{ 4h }{ 3 }+6\frac{ 2h }{ 3 } \right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[=\pi \left( 6\frac{ 2h }{ 3 } \right)\left( 62h \right)\] i think now you can complete

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first bracket gives h=9,rejected as h<9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0only value is h=3 find second derivative and check for maxima

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then find r and finally V by above formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you solve further?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0may i ask how you get the factor on the dV/dh?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you let out the common factor?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1using the 2nd D is a waste of time...you don't need it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4h }{ 3 }\left( 6\frac{ 2h }{ 3 } \right)+\left( 6\frac{ 2h }{ 3 } \right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now take the common factor out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have to show it is maximum or minimum value. Here by chance you get only one value,you may get more values.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you need to look at h=0,3,9 (end poins and critical values) V(0)=V(9)=0 V(3) will be the absolute max since it is positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when h=0 ,V=0 when h=9,r=0 V=0 actually cylinder with zero radius or zero height so no cylinder with these dimensions that is why 0<h<9

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you can use 0 and 9...you get a degenerate cylinder

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1the math works out. you don't need the 2nd D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it thank you very much
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