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Voluem of cylinder is pi * r^2 * h

Obviously, we can get the largest value if r or h are two of the largest value used.

Let's try r = 12, h = 10, v = pi*36*10 = 360pi
And let's try r = 10, h = 12, v = pi*25*12 = 300pi.

The largest possible is therefore 360pi cubic-inches
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I am getting maximum area as 200 pi..with r = 5 and h = 8
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My ans is h=8, r=5, volume=200pi.

(If h=12, r can't be more than 4. So pi*r^2*h=12*16pi=192pi
Apparently h=10, r=4 can't be more

Since vol. involves r^2, looking for a greater r would help. However r can't be 6 because the max possible length of the other side could be only 10 which could not fit a circular base)
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Remember: the base of the cylinder is a circle
r is constant around the circle, so the base can completely fit in a square, not a rectangle. Therefore, you will not be able to use the entire space in the base of the rectangular box.

the volume of a cylinder is pi * r^2 * h

select largest possible r such that the dimensions of the rectangular base allow maximum value or r
you would want to maximize r more than h because r is squared here
so, select bases 10 and 12 allowing a maximum value of r = 5

volume of cylinder: 25 pi * 8 = 200 pi

to check, think if 8 is one of the dimensions of the base, r = 4
volume of cyclinder: 16 pi * (either 10 or 12 ) = 160 pi or 192 pi
not maximum volume
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max vol answer is 200pi with r = 5, and height = 8

other are clearly lower,
r = 4, h = 12
r = 4, h = 10

u have to take care of the different sides of the box while calculating the volume...

to ncprasad's concern, a diagonal long cylinder can't be fit into the box unless it becomes a think stick...
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12 by 10, determines the radius of the cylinder = 10/2 = 5
So height 8
Ans 200pi
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yes the OA is 200pi... but i got trapped along with ywilfred by maximizing the Volume algebratically... and the trap was logical sense :evil:
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Why cant we have radius =6 and height = 10
making the volume 360 Pi ?
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Bunuel
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Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution:
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\);
If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\);
If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

So, the maximum volume is for \(200\pi\).


Similar question to practice: the-inside-dimensions-of-a-rectangular-wooden-box-are-128053.html

Hope it helps.
Hi, Can you explain why the diameter cannot be 12 ?I am not getting the concept clearly...Thanks
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farukqmul
Bunuel
Ashamock
Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution:
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\);
If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\);
If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

So, the maximum volume is for \(200\pi\).


Similar question to practice: the-inside-dimensions-of-a-rectangular-wooden-box-are-128053.html

Hope it helps.
Hi, Can you explain why the diameter cannot be 12 ?I am not getting the concept clearly...Thanks

Sure. If the diameter is 12 then it won't fit on any face of the box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Hope it's clear.
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willget800
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie
B. 200 Pie
C. 300 Pie
D. 320 Pie
E. 450 Pie

Always remember, the largest possible value will have a diameter that will need to feet on 2 of the sides. Therefore, if it fits in 10 it will also fit in 12. So choose a diameter of 10 = 2r, r=5 and then use the other side 8 for the height giving a total volume of 200pi
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willget800
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie
B. 200 Pie
C. 300 Pie
D. 320 Pie
E. 450 Pie

So first we need to place the cylinder so that its diameter is >= to at least two of the dimensions.
That is 10 and 12, so r = 5

Next, height has to be 8 then. I suggest you draw the cuboid and see for yourself that we need a side where the circumference stands to be 10*12 otherwise diameter won't fit

Then I guess the question asks for the largest volume possible

So pi (r ) ^ 2 * h = 200pi

That would be answer B indeed

Hope it helps
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willget800
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie
B. 200 Pie
C. 300 Pie
D. 320 Pie
E. 450 Pie
Question needs to be edited to show that we need to calculate volume.
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willget800
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?

A. 180 pie
B. 200 Pie
C. 300 Pie
D. 320 Pie
E. 450 Pie


if you take 12*10 as cylindrical base, you can occupy largest volume with the cylinder
so volume of the cylinder is : pi*5*5*8=200*pi
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Bunuel
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Why cant we have radius =6 and height = 10 making the volume 360 Pi ?

Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.

Complete solution:
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?

\(volume_{cylinder}=\pi{r^2}h\)

If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\);
If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\);
If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);

So, the maximum volume is for \(200\pi\).

Answer: B.

Similar question to practice: the-inside-dimensions-of-a-rectangular-wooden-box-are-128053.html

Hope it helps.
buunel this is a great explanation but it is hard to visualize without picture, can you draw pictures according to your dimension plz
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a doubt on basics :
it is a very silly question.
But why are we considering volume and not surface area ?
I got confused and made a blunder in the problem
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