A circular cylinder is to fit inside a sphere of radius 10.Calculate the maximum possible volume of the cylind

Question

(it is probably best to take as your independent variable the height of the cylinder)
Who can help me with this maths problem?

Answer 1

This is a good example of optimization and modelling. The first step is to draw a picture of the problem. It makes it easier to see what you're doing.



Secondly, we must determine what values we know, and what equations we can form using the information given. We know that the radius of the sphere is 10, and that the volume of the cylinder (V) is pi*(r^2)*h, where h is the height of the cylinder. We also know how the radius is related to the height and radius of the sphere.



radius of sphere: 10

V = pi * r^2 * h



In order to keep going, we need to determine how the radius of the cylinder is related to the height. We know that the centre of the cylinder will be located in the centre of the sphere, and that the top and bottom corners of the cylinder will touch the outside of the sphere. Therefore, we can can use Pythagoreans Theorem to solve for r. The triangle we will use is the radius and half the height as the legs and the radius of the sphere as the hypotenuse.



r^2 + (h/2)^2 = 10^2

Therefore... r^2 = 10^2 - (h/2)^2

Notice that we'll never need to know the value of r, since we use r^2 in our equation for volume.



Let's plug this new radius into our original volume equation.

V = pi * r^2 * h = pi * (100 - (h^2)/4) * h



Simplify... V = 100*pi*h - (pi/4)*h^3



In order to find the maximum of this equation, we need to find the derivative (dV/dh). Our function (and therefore, the volume of the cylinder) will be maximized at a critical point, where the derivative equals zero. Use the power rule to find the derivative.



dV/dh = 100*pi - (3*pi/4)*h^2 = 0

Solve for h... h^2 = (100*pi) / (3*pi/4)

Pi cancels out... h = sqrt(400/3) = 11.547



Now that we have the height, let's plug it back in for our volume equation...

V = 100*pi*h - (pi/4)*h^3 = 100*pi*11.547 - (pi/4)*11.547^3

So, our maximum volume is 2418.

Answer 2

I will first derive a general formula for this problem for you.

First you know that the volume of a cylinder is V=pi*r^2*h, we will assume the system is on a coordinate axis with the origin at the center of the sphere. Then we have V=pi*r^2*2h if h is the height from the center to the top of the cylinder. Now we must substitute in for h, so the Radius of the sphere (R)makes a right triangle with the height and the radius of the cylinder (r). This would make the height equal sqrt(R^2-r^2) using the pythagorean theorem.

Now the equation of volume is V(r)=pi*r^2*sqrt(R^2-r^2).

Take the derivative: V'(r)=4*pi*sqrt(R^2-r^2)+(R^2-r^2)^(-1/2)(-2r)=0

Solve for r in terms of r (skipping steps a bit here) : 2*r^3=4r*(R^2-r^2)

r=sqrt(2/3*R^2)=R*sqrt(2/3)

...Now you can plug in r=sqrt(2/3*R^2) for the volume equation(V(r)=pi*r^2*sqrt(R^2-r^2)) and put 10 in for R to find the maximum volume.



V(R)=pi*4/3*R^2*sqrt(1/3R^2)

V(10)=pi*4/3*100*sqrt(100/3)...

I'm pretty sure I've checked my work, but if anything is messed up (happens to me a lot), at least a general idea is there. Also, there might be (and probably is) a faster way to solve this.

Answer 3

First of all you have to know the volume of a cylinder, it is

V = pi(r^2)h WHERE r is the radius of the cylinder and h is the height.

Inside a sphere, there is a relationship of h, r, and the radius of the sphere we let a=10; now let's find it: inscribing such that the sphere is centred at the origin, then let the cylinder have its axis along the x axis. Draw this out.

Then, (h/2)^2 + r^2 = a^2 because the cylinder is touching the sphere at the radius a. This is the Pythagorean theorum, I hope that you can see this.

Next, I solved the above equation for h = 2[a^2-r^2]^(1/2)



The volume of the cylinder v = pi(r^2)h = 2pi(r^2)[a^2-r^2]

Divide both sides by 2pi and take the derivative of V with respect to r:

(1/2pi)dV/dr = 2r[a^2-r^2]^(1/2) - (r^3)/[a^2-r^2]^(1/2) set = 0

Solve the middle part of the above equal to 0 to gain the volume's maximum for r

and get r=a(2/3)^(1/2)

Then just plug in r into the formula for h and you have all the parameters necessary to put that damn cylinder into the sphere.

Answer 4

The volume of the sphere is found by following



y=4/3(pi)10^3



However, it'll be easier to work it out by using a hemisphere, so

y=2/3(pi)1000.



The volume of the cylinder in the hemisphere is found by



max = h(pi)r^2



where height and the radius of the cylinder is related in a way that

10^2 = height^2 + radius^2.



So let's say that the height of the cylinder is the independent value. Height of the cylinder is X, and the radius of the cylinder is (100-x^2)^(1/2).



Then we have the equation



yMAX = x(pi)((100-x^2)^(1/2))



Now we find the derivative of it.



dy/dx = (pi) [(1)(100-x^2)^(1/2) + (x)(1/2)(-2x)(100-x^2)^(-1/2)]



Which equals



dy/dx = (pi)(100-x^2)^(1/2)+(pi)(-x^2)(100-x^2)^(-1/2)



Now we set them equal to 0 and the pi's drop out



0 = (100-x^2)^(1/2)+1/[(-x^2)(100-x^2)^(1/2)]



Multiply both sides by (100-x^2)^(1/2)(-x^2)



Which gets you



(100-x^2)(-x^2)+1 = 0



-100x^2 + x^4 = -1



x^4 - 100x^2 + 1 = 0



That's as far as I can solve without a calculator or some kind of a textbook. Sorry, hope it helped (heck, I hope it's right). You'd probably want to go over the numbers.



Whatever answer you get though, remember that we're working with a hemisphere, and you should double the volume of the cylinder that you find.

Answer 5

the quantity of the cylinder is h*pi*r^2, the place h is the top of the cylinder and r is the radius of the tip of the cylinder. by way of fact the perimeters of the cylinder are touching the sector (2 small circles, like circles of selection on a globe), a diagonal of the cylinder, during the centre, is the comparable as a diameter of the sector and has length 2*R (R is the radius of the sector). Take 0.5 of all and sundry diagonal (they're each and all of the comparable). complete the corresponding triangle employing the 0.5-top of the cylinder and the radius of the cylinder. you at present have a superb-attitude triangle the place the hypothenuse is R and the two sides are h/2 and r. employing that (and the undeniable fact that R = 10), you have r^2 + (h/2)^2 = 10^2 exhibit r in terms of h (or h in terms of r). as an occasion r^2 = one hundred - (h/2)^2 you may then write an equation for the quantity of the cylinder employing only one variable (enable us to assert h): volume = h*pi*r^2 will become volume = h*pi*(one hundred-(h/2)^2) as quickly as you multiply by using, you have a polynomial in h for the quantity differentiate (dV/dh) and locate the value(s) of h the place dV/dh = 0. have exciting. careful with (h/2)^2 = (h^2)/4 no longer (h^2)/2