- Thread starter
- #1

- Thread starter leprofece
- Start date

- Thread starter
- #1

- Admin
- #2

2.) Can you state the constraint on the objective function? How will the cone being inscribed within a sphere constrain the objective function? What would a convenient set of coordinate axes be for this problem?

- Thread starter
- #3

ok lateral surface = pi R L And L = sqrt ( r +h

2.) Can you state the constraint on the objective function? How will the cone being inscribed within a sphere constrain the objective function? What would a convenient set of coordinate axes be for this problem?

Then applying thales I got sqrt ( h -r/r = H -R/R

I try to get r into the area formula, derive but i got a very large equation respect to h and I dont get the answer given

I have asked so to get if it is possible an easier way to get the answer maybe i am figuring out the problem bad

- Admin
- #4

If this is the case, that you are supposed to maximize the lateral surface area of the cone, I suggest looking at a cross-section of the two objects and see if you can express the radius $r$ of the cone in terms of the radius $R$ of the sphere and the height $h$ of the cone.

- Thread starter
- #5

- Admin
- #6

I was reading that the cone must be inscribed within a sphere, because you stated the cone was confined. And so I find the lateral surface area of the cone is maximized when the height of the cone is 4/3 times the radius of the sphere, and minimized at h=0 and h = 2R, as you can see from this plot (where I have let R=1):no i want the minimun

However, had I really been paying attention before, I would have noticed that given the stated answer, the sphere must be in fact circumscribed by the cone.

So, if we consider the cross-section of the two objects through their centers, there the center of the base of the cone is at the origin of our coordinate system, we find the slant height of the cone will lie along the line:

\(\displaystyle y=-\frac{h}{r}x+h\)

Now, this slant height will be tangent to the circle (representing the sphere of radius $R$):

\(\displaystyle x^2+(y-R)^2=R^2\)

So, what we may do is then substitute for $y$, and then require the discriminant of the resulting quadratic in $x$ to be zero. You should be able to use this to show:

\(\displaystyle r^2=\frac{hR^2}{h-2R}\)

Substituting this into the objective function (the lateral surface $S$ of the cone) you should be able to show that:

\(\displaystyle S(h)=\frac{\pi hR(h-R)}{h-2R}\)

Differentiating this with respect to $h$, and equating the result to zero, you should find this implies:

\(\displaystyle h^2-4Rh+2R^2=0\)

And from this, discarding the negative root because we require $2R<h$, we then find:

\(\displaystyle h=\left(2+\sqrt{2} \right)R\)

Now, at each step where I state "you should be able to show..." this is your cue to do just that. I have merely provided the outline of the steps that can be taken to get the desired result. I encourage you to work through the algebra and calculus and will be happy to assist if you get stuck, as I have worked out each step. I would just ask that you show what you did leading up to where you are stuck if this happens.

- Thread starter
- #7

Ok I dont understand only How you get the lineI was reading that the cone must be inscribed within a sphere, because you stated the cone was confined. And so I find the lateral surface area of the cone is maximized when the height of the cone is 4/3 times the radius of the sphere, and minimized at h=0 and h = 2R, as you can see from this plot (where I have let R=1):

View attachment 1886

However, had I really been paying attention before, I would have noticed that given the stated answer, the sphere must be in fact circumscribed by the cone.

So, if we consider the cross-section of the two objects through their centers, there the center of the base of the cone is at the origin of our coordinate system, we find the slant height of the cone will lie along the line:

\(\displaystyle y=-\frac{h}{r}x+h\)

Now, this slant height will be tangent to the circle (representing the sphere of radius $R$):

\(\displaystyle x^2+(y-R)^2=R^2\)

So, what we may do is then substitute for $y$, and then require the discriminant of the resulting quadratic in $x$ to be zero. You should be able to use this to show:

\(\displaystyle r^2=\frac{hR^2}{h-2R}\)

Substituting this into the objective function (the lateral surface $S$ of the cone) you should be able to show that:

\(\displaystyle S(h)=\frac{\pi hR(h-R)}{h-2R}\)

Differentiating this with respect to $h$, and equating the result to zero, you should find this implies:

\(\displaystyle h^2-4Rh+2R^2=0\)

And from this, discarding the negative root because we require $2R<h$, we then find:

\(\displaystyle h=\left(2+\sqrt{2} \right)R\)

Now, at each step where I state "you should be able to show..." this is your cue to do just that. I have merely provided the outline of the steps that can be taken to get the desired result. I encourage you to work through the algebra and calculus and will be happy to assist if you get stuck, as I have worked out each step. I would just ask that you show what you did leading up to where you are stuck if this happens.

y = -x/r +h

I have talked to you from tales

any way it was a very good answer

if you dont mind canyou explain me how do you get the line??'

CESAr

- Admin
- #8

We see the line passes through the points $(0,h)$ and $(r,0)$. One easy method we may use to get the equation of the line is to use the two-intercept formula:

\(\displaystyle \frac{x}{r}+\frac{y}{h}=1\)

Now, solving for $y$, we obtain:

\(\displaystyle y=-\frac{h}{r}x+h\)

You could also determine the slope of the line, and the use either point in the point-slope formula.

- Thread starter
- #9

all right I understood everything but now i am stuck is when \(\displaystyle x^2+(y-R)^2=R^2\)I was reading that the cone must be inscribed within a sphere, because you stated the cone was confined. And so I find the lateral surface area of the cone is maximized when the height of the cone is 4/3 times the radius of the sphere, and minimized at h=0 and h = 2R, as you can see from this plot (where I have let R=1):

View attachment 1886

However, had I really been paying attention before, I would have noticed that given the stated answer, the sphere must be in fact circumscribed by the cone.

So, if we consider the cross-section of the two objects through their centers, there the center of the base of the cone is at the origin of our coordinate system, we find the slant height of the cone will lie along the line:

\(\displaystyle y=-\frac{h}{r}x+h\)

Now, this slant height will be tangent to the circle (representing the sphere of radius $R$):

\(\displaystyle x^2+(y-R)^2=R^2\)

So, what we may do is then substitute for $y$, and then require the discriminant of the resulting quadratic in $x$ to be zero. You should be able to use this to show:

\(\displaystyle r^2=\frac{hR^2}{h-2R}\)

Substituting this into the objective function (the lateral surface $S$ of the cone) you should be able to show that:

\(\displaystyle S(h)=\frac{\pi hR(h-R)}{h-2R}\)

Differentiating this with respect to $h$, and equating the result to zero, you should find this implies:

\(\displaystyle h^2-4Rh+2R^2=0\)

And from this, discarding the negative root because we require $2R<h$, we then find:

\(\displaystyle h=\left(2+\sqrt{2} \right)R\)

Now, at each step where I state "you should be able to show..." this is your cue to do just that. I have merely provided the outline of the steps that can be taken to get the desired result. I encourage you to work through the algebra and calculus and will be happy to assist if you get stuck, as I have worked out each step. I would just ask that you show what you did leading up to where you are stuck if this happens.

Ok i substitute X^2+y^2 -2Ry+R^2 = R^2

Cancel R and substiute x+(H+hx/r)^2 +2R(H+hx/r)

Apllying algebra it remains a equation and it is impossible for me to get r= HR/H-2R

Could you please tell me how to get there ??

- Admin
- #10

\(\displaystyle x^2+\left(-\frac{h}{r}x+h-R \right)^2=R^2\)

and after you expand and combine like terms, you should find you can arrange the result as a quadratic in $x$ in standard form:

\(\displaystyle \left(h^2+r^2 \right)x^2+2hr(R-h)x+hr^2\left(h-2R \right)=0\)

Make sure you can do this on your own, as being able to do such algebra is a very important skill for analytic geometry. I want to give you a chance to do this on your own, but if after trying you find that you just can't get there, I will show you step by step how to do it, but I want to see what you tried and where you are stuck so I can better help you.

Once you get to this point, you then want to equate the discriminant to zero (since the line and the circle are tangent), and this will allow you to represent $r^2$ in terms of the other parameters.