Simultaneous equations and the volume of a cylinder

[solo]

Hi everyone, I've started CIE P1 and there's a question on simultaneous equations and working out the volume. The first part asked to make a simultaneous equation, which I have as the following:

4r+h=100
2pirh+2pir^2=1400pi

It then asks to work out the volume in relation to pi, saying that there are two answers. Using the substitution method of isolating h and then putting this in to the quadratic equation I got:

(2r-20)(3r-70)=0

From this there are two answers. When r=10, h= 60 or
r= 70/3, h= 100-(280/3).

It asks for the volume of the sphere in relation to pi. I got the answer of 6000 pi when r= 10 but the book also has the answer 98000pi/27 cm^3, which is assume is when r= 70/3 but I'm not sure.

Any help would be appreciated! Thanks

 

[stapel]

Hi everyone, I've started CIE P1 and there's a question on simultaneous equations and working out the volume. The first part asked to make a simultaneous equation, which I have as the following:

4r+h=100
2pirh+2pir^2=1400pi

What was the original question?

...It asks for the volume of the sphere in relation to pi. I got the answer of 6000 pi when r= 10 but the book also has the answer 98000pi/27 cm^3, which is assume is when r= 70/3 but I'm not sure.

Why are you not sure? What did you get when you plugged this r-value into the formula for the volume of a sphere?

Please be complete. Thank you!

 

[solo]

Thanks for the response. I made a mistake in my thread and it should have said it asks for the volume of the cylinder and not sphere.

Here's the original question:

The diagram shows the net of a cylindrical container of radius r cm and height
h. The full width of the metal sheet from which the container is made is 1 m,
and the shaded area is waste. The surface area of the container is 1400 cm^2.

(ii Write down a pair of simultaneous equations for rand h.
(i i) Find the volume of the container, giving your answers in terms pi.
(There are two possible answers.)


The first answer I got was 6000pi which came from the first value of r=10. The answer at the back says 6000pi or 98000pi/27 cm^2. I'm unsure how this answer came about and assumed it was from following the same process with the other value of r but I may be mistaken.

This is what I got for the other value of r. I started from the answer that I got from the quadratic, 3r-70 =0. From that r= 70/3 putting this part into 4r+ h = 100 which gives the answer h= 100-(280/3) which is approx 6.67. So I have r and h and put it in the formula for a cylinder which gives 11375.95cm ^3, which then divided by pi is roughly 3621.1 pi.

I haven't studied maths for a long time so it may be rustiness and something really obvious. Thanks.

 

[solo]

Do the replies normally take a while? I responded but nothing seems to have updated?

 

[stapel]

Do the replies normally take a while? I responded but nothing seems to have updated?

I'm sorry, but whoever told you that there are paid staff-members waiting on-hand to provide instant replies was very much mistaken. Instead, what you read in the "Read Before Posting" announcement continues to hold; namely, this forum is "staffed", if I can use that term, by volunteers who surf by as they are able. If you're needing on-demand responses, you'll want to consider contracting with a service which offers such. Apologies for the confusion.

 

[stapel]

Here's the original question:

The diagram shows the net of a cylindrical container of radius r cm and height h. The full width of the metal sheet from which the container is made is 1 m, and the shaded area is waste. The surface area of the container is 1400 cm^2.

(i) Write down a pair of simultaneous equations for r and h.
(ii) Find the volume of the container, giving your answers in terms pi. (There are two possible answers.)

Thank you! However, we are unable to see "the diagram" or "the shaded area". Also, are we supposed to be using the "waste", or is this stuff that doesn't count?

(i) The first answer I got was 6000pi which came from the first value of r=10. My working was:

. . .4r + h = 100

How did you obtain this equation? In particular, what is the source of the "100"?

. . .2 pi r h + 2 pi r^2 = 1400 pi

I will guess that this was derived as follows:

. . .The formula for the surface area SA of a cylinder
. . .with radius r and height h is:

. . . . .SA = 2(pi r^2) + (2 pi r)h

. . .We are given that SA = 1,400 pi square units, so:

. . . . .2 pi r^2 + 2 pi r h = 1,400 pi

If so, then I agree with your work and reasoning.

(ii) For this part, I started by using substitution:

. . .(2r - 20) (3r - 70) = 0

What did you substitute? Into where?

Assuming your first equation to be correct, I will guess that your work and reasoning was along these lines:

. . . . .4r + h = 100

. . .Then:

. . . . .h = 100 - 4r

. . .Plugging this in for "h" in the second equation, we get:

. . . . .2 pi r^2 + 2 pi r (100 - 4r) = 1,400 pi

. . .Dividing through by 2pi, we get:

. . . . .r^2 + r(100 - 4r) = 700

. . . . .r^2 + 100r - 4r^2 = 700

. . . . .100r - 3r^2 = 700

. . . . .0 = 3r^2 - 100r + 700

. . . . .0 = (3r - 70)(r - 10)

. . . . .r = 70/3 or r = 10

If so, then I agree with your work and reasoning.

From this there are two answers: r = 10 or r = 70/3. When r=10, h= 60; when r= 70/3, h= 100-(280/3).

Yes, but it would probably be better (easier for you, plus it's the answer the grader is expecting) to simplify the fractional form to 20/3.

This is what I got for the other value of r. I started from the answer that I got from the quadratic, 3r-70 =0. From that r= 70/3 putting this part into 4r+ h = 100 which gives the answer h= 100-(280/3) which is approx 6.67.

I'm sorry, but I'm not sure what you're doing here...? (By the way, pretty much always you should use the "exact" form, rather than the decimal approximation, at least until the very end.)

The answer at the back says 6000pi or 98000pi/27 cm^2. I'm unsure how this answer came about and assumed it was from following the same process with the other value of r but I may be mistaken.

They plugged the exact forms of r and h (clearly, the forms they're wanting) into the volume formula.

So I have r and h and put it in the formula for a cylinder which gives 11375.95cm ^3, which then divided by pi is roughly 3621.1 pi.

What did you get, when you used the exact form in the volume formula?

. . .V = pi r^2 h

. . . . .= pi (70/3)^2 (20/3)

. . . . .= pi (4900/9) (20/3)

. . . . .= pi [(4900 * 20) / (9 * 3)]

. . . . .= ....

...and so forth.

 

[solo]

Thanks for the reply, I did read the rules but can be a bit absent with details at times, seems to have affected my forum posts as well as my maths! I'll take that in to consideration next time.

You're correct with everything you have said and it seems I should have gone and tried the exact form. Thanks for the help, will try and play around a bit more before panicking that I don't know.