jarp0120
  • jarp0120
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
what have you tried so far ?
Jhannybean
  • Jhannybean
Looks like an optimization problem
jarp0120
  • jarp0120
i've tried to solve it but i don't know what to do and also how to insert the derivatives

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jarp0120
  • jarp0120
they have the same SA right?
ganeshie8
  • ganeshie8
How so ? Maybe, as a start draw a nice sketch of cylinder inside a cone and label the given dimensions
ganeshie8
  • ganeshie8
I think, this problem is equivalent to finding the volume of the largest cylinder that can be inscribed inside a cone..
jarp0120
  • jarp0120
ohw now i get the figure
jarp0120
  • jarp0120
|dw:1443926956736:dw| like this?
Jhannybean
  • Jhannybean
|dw:1443926942036:dw|
Jhannybean
  • Jhannybean
Yeah there you go lol.
ganeshie8
  • ganeshie8
That looks good! label the given dimensions also..
ganeshie8
  • ganeshie8
show radius 6 and height 9 in the diagram
anonymous
  • anonymous
|dw:1443937323135:dw| volume of cylinder \[V = \pi r^2 h\]
Jhannybean
  • Jhannybean
Naice.
jarp0120
  • jarp0120
|dw:1443927181720:dw|
anonymous
  • anonymous
\[V=\pi \left( 6-\frac{ 2h }{ 3 } \right)^2 h\] diff. and find stationary values
anonymous
  • anonymous
and then find second derivative and see where it is maxima
Zarkon
  • Zarkon
it is a constrained min/max problem. plug stationary points and boundary points into volume formula
Zarkon
  • Zarkon
pick the largest
Zarkon
  • Zarkon
\[0\le h\le 9\]
jarp0120
  • jarp0120
should simplify it? i'm confused
jarp0120
  • jarp0120
i don't know what to do with the volume part :(
anonymous
  • anonymous
\[\frac{ dV }{ dh }=\pi[h*2(6-\frac{ 2h }{ 3 })*\frac{ -2 }{ 3 }+1\left( 6-\frac{ 2h }{ 3 } \right)^2]\] simplify and put dV/dh =0 and find values of h
jarp0120
  • jarp0120
okay
anonymous
  • anonymous
\[=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left[ \frac{ -4h }{ 3 }+6-\frac{ 2h }{ 3 } \right]\]
anonymous
  • anonymous
remember 0
anonymous
  • anonymous
\[=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left( 6-2h \right)\] i think now you can complete
anonymous
  • anonymous
first bracket gives h=9,rejected as h<9
anonymous
  • anonymous
only value is h=3 find second derivative and check for maxima
anonymous
  • anonymous
then find r and finally V by above formula.
anonymous
  • anonymous
can you solve further?
jarp0120
  • jarp0120
may i ask how you get the factor on the dV/dh?
jarp0120
  • jarp0120
you let out the common factor?
Zarkon
  • Zarkon
using the 2nd D is a waste of time...you don't need it
anonymous
  • anonymous
\[-\frac{ 4h }{ 3 }\left( 6-\frac{ 2h }{ 3 } \right)+\left( 6-\frac{ 2h }{ 3 } \right)^2\]
anonymous
  • anonymous
now take the common factor out
anonymous
  • anonymous
you have to show it is maximum or minimum value. Here by chance you get only one value,you may get more values.
Zarkon
  • Zarkon
you 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
  • anonymous
when 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
Zarkon
  • Zarkon
you can use 0 and 9...you get a degenerate cylinder
Zarkon
  • Zarkon
the math works out. you don't need the 2nd D
jarp0120
  • jarp0120
H=3 Then r=4 V=48pi
jarp0120
  • jarp0120
I got it thank you very much

Looking for something else?

Not the answer you are looking for? Search for more explanations.