Need some help with the 2006 Paper 2 Question 12 b

Thanks!

**0**

# 2006 Paper 2 Question 12 b

Started by Philip, May 20 2009 10:43 AM

2 replies to this topic

### #1

Posted 20 May 2009 - 10:43 AM

Yeah!

### #2

Posted 20 May 2009 - 11:24 AM

QUOTE (Philip @ May 20 2009, 11:43 AM) <{POST_SNAPBACK}>

Need some help with the 2006 Paper 2 Question 12 b

Thanks!

Thanks!

Handily you can find the HSN worked solution at Google Books but here's how I would do it (not too different really)...

Remember that extremes (that is the mins and maxes) can occur at stationary points or

__the start and end of the given range__(underlined as that's probably what you've done wrong )

So, A = 80 - 12

*x*- 48^-1

dA/d

*x*= -12 + 48

*x*^-2 = 0 (for stat. point)

-12 + 48/

*x*^2 = 0

48/

*x*^2 = 12

12

*x*^2 = 48

*x*^2 = 4

*x*= 2 or -2

Thus

*x*= 2 as

*x*> 0.

So find A for the

__values of x...__

**three**When

*x*= 1, A = 80 - 12x1 - 48/1^2 = 20

When

*x*= 2, A = 80 - 12x2 - 48/2^2 = 32

When

*x*= 4, A = 80 - 12x4 - 48/4^2 = 20

Therefore: Max of A is 32 when

*x*= 2 and Min of A is 20 when

*x*= 1 or 4

Beware negatives,

for they will haunt you in haikus

and in exam rooms.

for they will haunt you in haikus

and in exam rooms.

### #3

Posted 20 May 2009 - 01:26 PM

QUOTE (Garden @ May 20 2009, 12:24 PM) <{POST_SNAPBACK}>

QUOTE (Philip @ May 20 2009, 11:43 AM) <{POST_SNAPBACK}>

Need some help with the 2006 Paper 2 Question 12 b

Thanks!

Thanks!

Handily you can find the HSN worked solution at Google Books but here's how I would do it (not too different really)...

Remember that extremes (that is the mins and maxes) can occur at stationary points or

__the start and end of the given range__(underlined as that's probably what you've done wrong )

So, A = 80 - 12

*x*- 48^-1

dA/d

*x*= -12 + 48

*x*^-2 = 0 (for stat. point)

-12 + 48/

*x*^2 = 0

48/

*x*^2 = 12

12

*x*^2 = 48

*x*^2 = 4

*x*= 2 or -2

Thus

*x*= 2 as

*x*> 0.

So find A for the

__values of x...__

**three**When

*x*= 1, A = 80 - 12x1 - 48/1^2 = 20

When

*x*= 2, A = 80 - 12x2 - 48/2^2 = 32

When

*x*= 4, A = 80 - 12x4 - 48/4^2 = 20

Therefore: Max of A is 32 when

*x*= 2 and Min of A is 20 when

*x*= 1 or 4

For the three values at the end: did you just use the x=1 and x=4 from the graph to find the mins and maxs?

What i did wrong was that i found the minimum x=2, but didn't know you just subbed in the other two values.

Thanks for the help

Yeah!

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