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Maximizing the amount of water displaced by a sphere inserted in a cone

leprofece

Member
Jan 23, 2014
241
choose the diameter of a sphere so that when it is inserted into a cope of form conic (depth H and RADIUS R) fill of water, spilling as much as possible of liquid when the sphere rests is on the walls of cope. ( volume of a segment spherical of radius "r" y height "h' es: V = pih2{r- ( h ))

Answer
r= HRL/(L-R)(L+2R)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

I would draw a diagram as follows:

leprofece4.jpg

From this you can get the constraint. For the objective function, we need the volume of a spherical cap:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

$a$ will be the positive $x$-intercept of the circle and $h=r+k$.

Can you proceed?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

I would draw a diagram as follows:

View attachment 2005

From this you can get the constraint. For the objective function, we need the volume of a spherical cap:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

$a$ will be the positive $x$-intercept of the circle and $h=r+k$.

Can you proceed?
it may be
V= pi(r-k)/6(3a2+(r-k)2)
i must derive may I Derive now?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

it may be
V= pi(r-k)/6(3a2+(r-k)2)
i must derive may I Derive now?
No...$h=r+k$, you need to find $a$ as I described (the positive $x$-intercept of the circle), and you need to find the constraint. The most straightforward way I can think of is to use the fact that a radius of the circle drawn to the point of tangency of the line segment from $(0,H)$ to $(R,0)$ is perpendicular to this segment, and so the formula for the perpendicular or shortest distance between a point and a ine may be used to obtain the value of $k$ in terms of $R$, $H$ and $r$. Another method which yields the same result, but requires more algebra is to equate the equation of the line to the circle and require the discriminant of the resulting quadratic to be zero.

The formula for the distance $d$ between a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$ is:

\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)

We know in this case that $d=r$. Can you identify $m$ and $b$?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

No...$h=r+k$, you need to find $a$ as I described (the positive $x$-intercept of the circle), and you need to find the constraint. The most straightforward way I can think of is to use the fact that a radius of the circle drawn to the point of tangency of the line segment from $(0,H)$ to $(R,0)$ is perpendicular to this segment, and so the formula for the perpendicular or shortest distance between a point and a ine may be used to obtain the value of $k$ in terms of $R$, $H$ and $r$. Another method which yields the same result, but requires more algebra is to equate the equation of the line to the circle and require the discriminant of the resulting quadratic to be zero.

The formula for the distance $d$ between a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$ is:

\(\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}\)

We know in this case that $d=r$. Can you identify $m$ and $b$?
Lets see
If the points are (0,H) y (R,0)
m must be -H/R
Now y = -Hx/R +HR/R so b = H
I dont know if L is K in your solution
Introducing in your formula
r = - -Hx/R +H/(sqrt) ( - (-H/R)^2 +1)
But I dont proceed because in the book answer it is L
There must be something wrong in my solution
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

I would assume your book is labeling:

\(\displaystyle L=\sqrt{R^2+H^2}\)
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

I would assume your book is labeling:

\(\displaystyle L=\sqrt{R^2+H^2}\)
I dont know if my calculation was good
Because I dont know how to pul this \(\displaystyle L=\sqrt{R^2+H^2}\)[/QUOTE]
in my given solving . ??????'
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

You need to determine $k$. Use the formula I gave for the distance between a point and a line where:

\(\displaystyle \left(x_0,y_0 \right)=(0,k)\)

\(\displaystyle m=-\frac{H}{R}\)

\(\displaystyle b=H\)

\(\displaystyle d=r\)

Do you see why this applies?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

You need to determine $k$. Use the formula I gave for the distance between a point and a line where:

\(\displaystyle \left(x_0,y_0 \right)=(0,k)\)

\(\displaystyle m=-\frac{H}{R}\)

\(\displaystyle b=H\)

\(\displaystyle d=r\)

Do you see why this applies?
The formula for the distance d between a point (x0,y0) and a line y=mx+b is:

d=∣∣mx0+b−y0∣∣m2+1−−−−−−√

We know in this case that d=r. Can you identify m and b?

so r= d = -H/R+H-k/(sqrt(h^2+r^2)/(R^2))

r= -H/R+H-k/((L)/(R))

So k = Lr+H/R-H/R

K = HR-Hx-rL/R

K=H-r

if I equalled and solved for r and i got nothing HR-Hx-rL/R = H-r



And now what must I do??
Derive that??)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

No, what you want to do is:

\(\displaystyle r=\frac{\left|-\frac{H}{R}\cdot0+H-k \right|}{\sqrt{\left(-\frac{H}{R} \right)^2+1}}\)

Now solve this for $k$ (observe that $H>k$) and substitute into the objective function. And you STILL need to find the positive $x$-intercept of the circle to use for $a$. Recall, your objective function is:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

and $h=r+k$. This is why you need $k$ and $a$.
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

No, what you want to do is:

\(\displaystyle r=\frac{\left|-\frac{H}{R}\cdot0+H-k \right|}{\sqrt{\left(-\frac{H}{R} \right)^2+1}}\)

Now solve this for $k$ (observe that $H>k$) and substitute into the objective function. And you STILL need to find the positive $x$-intercept of the circle to use for $a$. Recall, your objective function is:

\(\displaystyle V=\frac{\pi h}{6}\left(3a^2+h^2 \right)\)

and $h=r+k$. This is why you need $k$ and $a$.
con x = 0
Then i got K = HR-Lr/R

and a2 = 2V/piH- H2/3

then??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 276

If you mean:

k = (HR - Lr)/R

then that is correct. However, $a$ is the positive $x$-intercept of the circle:

\(\displaystyle x^2+(y-k)^2=r^2\)

Hence:

\(\displaystyle a=\sqrt{r^2-k^2}\)

Now, put the value of $k$ in there...and try to simplify.

Now, after you do this, what do you suppose you should do with these values?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 276

If you mean:

k = (HR - Lr)/R

then that is correct. However, $a$ is the positive $x$-intercept of the circle:

\(\displaystyle x^2+(y-k)^2=r^2\)

Hence:

\(\displaystyle a=\sqrt{r^2-k^2}\)

Now, put the value of $k$ in there...and try to simplify.

Now, after you do this, what do you suppose you should do with these values?
2V/piH- H2/3 =r2-k2

2V/piH- H2/3 =r2-(HR-Lr)2/R2

2V/piH- H2/3 =r2 - ( HR)2-2HRLr+(Lr)2/R2