# Urgent AS Maths help please ! Watch

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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.

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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#2

(Original post by

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.

**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.

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#3

**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.

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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#4

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.

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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(Original post by

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.

**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.

2Pi r h + 2Pi x r^2

So

2pi x r x 12 + 2pi x r^2 = 128pi ??

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#6

(Original post by

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 ??

**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 ??

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#7

(Original post by

I can tell this is wrong but...

**Aty100**)I can tell this is wrong but...

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(Original post by

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.

**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.

Thanks once again for you help !

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(Original post by

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.

**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.

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**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.

r = -16 & r = 4

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

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#11

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.

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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