Help appreciated (1 Viewer)

mattstonestreet

New Member
Joined
Jul 22, 2005
Messages
19
Gender
Male
HSC
2006
Just going throught the '97 maths paper and em having difficulty with 6iii. If you do the question you can see that this is part of the simplification after differentiation. So basically I need to simplify this. If you write it out it isn't as bad as it looks here!

1/2(25-10x)^1/2 + 1/2x((2(25-10x)^1/2)^-1 =0
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
Let A=1/2.x(25-10x)1/2

dA/dx=1/2.{sqrt(25-10x) - 5x/sqrt(25-10x)} by product and chain rule

When dA/dx=0,

1/2.{sqrt(25-10x) - 5x/sqrt(25-10x)}=0

sqrt(25-10x) = 5x/sqrt(25-10x)

[sqrt(25-10x)][sqrt(25-10x)]=5x

25-10x=5x

15x=25

x=5/3

Then substitute this into A to find maximum area. You should also show that it's a max at this x value either using first or second derivative. Hope that helps.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top