Aty100
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I really need some help on this question, i would really appreciate if someone could guide me through the steps of what to do. Thanks in advance.

Question

A sealed metal container for food is a cyclinder of height 12cm and base radius r cm. Given the surface area of the container must be at most 128pi cm^2.

a) show that r^2 + 12r - 64 </= 0

b) hence find the maximum value of r.
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TeeEm
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(Original post by Aty100)
I really need some help on this question, i would really appreciate if someone could guide me through the steps of what to do. Thanks in advance.

Question

A sealed metal container for food is a cyclinder of height 12cm and base radius r cm. Given the surface area of the container must be at most 128pi cm^2.

a) show that r^2 + 12r - 64 </= 0

b) hence find the maximum value of r.
look at similar examples here
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math42
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(Original post by Aty100)
I really need some help on this question, i would really appreciate if someone could guide me through the steps of what to do. Thanks in advance.

Question

A sealed metal container for food is a cyclinder of height 12cm and base radius r cm. Given the surface area of the container must be at most 128pi cm^2.

a) show that r^2 + 12r - 64 </= 0

b) hence find the maximum value of r.
a) Do you know the formula for surface area of a cylinder? The circle ends should be apparent; as for the rest, you essentially have a lot of circumferences making up a certain height,

b) Think about the graph of r^2 + 12r - 64. It will have two roots, inbetween which it is less than zero.
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Pangol
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Part (a) should be pretty simple. Just use the equation for the surface area of a cylinder. Look it up if you have to, but it should be quite easy to work out - the top and bottom are just circles, and I'm sure that you know the equation for the area of a circle, and the vertical outside can be cut open to reveal a rectangle of height 12 and base... well, you tell me!

Part (b) is a quadratic inequality, and I suspect it is this that you are really being tested on. I'd go through this process;

- replace the inequality with an equality, so that you have a quadratic equation.
- solve this equation to get two solutions (probably)
- sketch the quadratic, using your two solutions as the intersections with the horizontal axis
- now look at the original inequality and see if the answer suggests itself.
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Aty100
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(Original post by 13 1 20 8 42)
a) Do you know the formula for surface area of a cylinder? The circle ends should be apparent; as for the rest, you essentially have a lot of circumferences making up a certain height,

b) Think about the graph of r^2 + 12r - 64. It will have two roots, inbetween which it is less than zero.
Yh so the formula for surface area of a cylinder is
2Pi r h + 2Pi x r^2

So

2pi x r x 12 + 2pi x r^2 = 128pi ??
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math42
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(Original post by Aty100)
Yh so the formula for surface area of a cylinder is
2Pi r h + 2Pi x r^2

So

2pi x r x 12 + 2pi x r^2 = 128pi ??
Well less than or equal to but yeah
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math42
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(Original post by Aty100)
I can tell this is wrong but...
I'm not sure what you're doing here. Just divide throughout by 2pi in the equation you had at first and rearrange and the answer follows.
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Aty100
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(Original post by 13 1 20 8 42)
I'm not sure what you're doing here. Just divide throughout by 2pi in the equation you had at first and rearrange and the answer follows.
Of course! I'm so stupid sometimes

Thanks once again for you help !
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Aty100
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(Original post by Pangol)
Part (a) should be pretty simple. Just use the equation for the surface area of a cylinder. Look it up if you have to, but it should be quite easy to work out - the top and bottom are just circles, and I'm sure that you know the equation for the area of a circle, and the vertical outside can be cut open to reveal a rectangle of height 12 and base... well, you tell me!

Part (b) is a quadratic inequality, and I suspect it is this that you are really being tested on. I'd go through this process;

- replace the inequality with an equality, so that you have a quadratic equation.
- solve this equation to get two solutions (probably)
- sketch the quadratic, using your two solutions as the intersections with the horizontal axis
- now look at the original inequality and see if the answer suggests itself.
Thanks so much
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Aty100
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(Original post by 13 1 20 8 42)
a) Do you know the formula for surface area of a cylinder? The circle ends should be apparent; as for the rest, you essentially have a lot of circumferences making up a certain height,

b) Think about the graph of r^2 + 12r - 64. It will have two roots, inbetween which it is less than zero.
So
r = -16 & r = 4

So would the answer for part b) be 4 ??
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Pangol
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Just realised that I didn't read part (b) properly, so my advice wasn't great. Well, it was, but for a slightly different question!

I think that the best thing to do is to take the original inequality, make it an equality, and complete the square. A sketch will still very much help you.
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math42
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(Original post by Aty100)
So
r = -16 & r = 4

So would the answer for part b) be 4 ??
Should be
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