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- #1

Answer R = H = 3 cm

V= pir^2h/3 cone

V Hemisfere = 2pir^3/3

I dont know how to solve h if in the other there is no H

- Thread starter leprofece
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- Thread starter
- #1

Answer R = H = 3 cm

V= pir^2h/3 cone

V Hemisfere = 2pir^3/3

I dont know how to solve h if in the other there is no H

- Feb 5, 2012

- 1,621

Hi leprofece,

Answer R = H = 3 cm

V= pir^2h/3 cone

V Hemisfere = 2pir^3/3

I dont know how to solve h if in the other there is no H

I am not sure I understand your problem. Did you translate this from another language?

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- #3

I believe the problem to be:Hi leprofece,

I am not sure I understand your problem. Did you translate this from another language?

A right circular cylinder capped with a hemisphere at one end (we'll call this object "the body") is enclosed by a right circular cone of radius 5 cm and height 15 cm, such that the base of the cone and the hemispherical cap are tangent to one another and the cone and "the body" share the same axis of symmetry. Here is a cross-section of the two bodies, through this common axis of symmetry:

Find the dimensions of "the body" which maximizes its volume.

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First, can you state the objective function? Then can you state the constraint using the cross section? I would approach the constraint using coordinate geometry.Yes that is right

it is correct yoour solving

may you help me??

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The volume of the body is the sum of that of a cylinder and a hemisphere.It is complicate

From the figure volume of the figure introduced

h (r^2) = V

It remains a triangle maybe it applies Thales

to get the another equation 2r/h = 2h/x

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- #8

ok it must be one of the two equationsThe volume of the body is the sum of that of a cylinder and a hemisphere.

pir

and the another must be by thales

????'

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- #9

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- #11

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- #12

y = hx/r is ready soConsider this diagram:

View attachment 2082

Now, can you find $h$ in terms of the constants $H$ and $R$ and the variable $r$?

the another part of the triangle is thales

h-r/R = H/r

So I must solve from here must not I ???

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- #13

That's not what I get. I used the fact that:y = hx/r is ready so

the another part of the triangle is thales

h-r/R = H/r

So I must solve from here must not I ???

\(\displaystyle a+h+r=H\)

and from similarity or the equation of the line, we find:

\(\displaystyle a=\frac{H}{R}r\)

Now, substitute for $a$ into the first equation, and solve for $h$. What do you find?

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- #14

I FindThat's not what I get. I used the fact that:

\(\displaystyle a+h+r=H\)

and from similarity or the equation of the line, we find:

\(\displaystyle a=\frac{H}{R}r\)

Now, substitute for $a$ into the first equation, and solve for $h$. What do you find?

Hr/R+r + h = H

Putting the data problem

5H/R+5+15 = H

H-5H/R = 20

HR -5H = 20 R

H(R-5) = 20 R

H=20R/R-5

derivating

I dont get H=R = 3 that is the book answer ·

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- #15

\(\displaystyle \frac{H}{R}r+h+r=H\)

Solving for $h$, we find:

\(\displaystyle h=H-r\left(1+\frac{H}{R} \right)\)

Now, our objective function is:

\(\displaystyle V(r,h)=\pi r^2h+\frac{2}{3}\pi r^3\)

Substituting for $h$, we obtain:

\(\displaystyle V(r)=\pi r^2\left(H-r\left(1+\frac{H}{R} \right) \right)+\frac{2}{3}\pi r^3\)

Distributing and combining like terms, we get:

\(\displaystyle V(r)=\pi Hr^2+\pi r^3\left(\frac{2}{3}-\left(1+\frac{H}{R} \right) \right)\)

This is the function you want to maximize.