The Edexcel C2 (24/05/12 - AM) Revision Thread

Ok. Last post of the night.
Capacity is volume.
It looks weird, but its a trapezium prism if such a thing exists.
Like all prisms, work out the Area of the shaped side x length
Area of a trapezium = (1/2)(B1 + B2)H
(1/2)(8x + 10x)(x)


Then times by length (y)

So Volume = (1/2)(8x + 10x)(x)(y)


and go from there.

(Hope I'm right :3)

I swear whoever thought logs were a good idea needs shooting. "ooh it'll make it simpler" make it simpler my arse.

LOL!

They're not that bad:P

Do you want the whore-of-this-thread award?

how would you do part b, and yes you're right as I checked the mark scheme but it doesn't explain how to do part b really well

How is she a whore? hahahaha get a grip k.

Do you want the whore-of-this-thread award?

So this is a really hard question that I'm stuck on, anyone care to help me. I can do part a but not the rest :/

Oh this is a fun question @_@

Part b is now focusing on surface area.
So the surface area of this tray = The two trapezium sides + the two rectangle sides and the large rectangle base.

Trapazoids
A=(1/2)(8x+10x)(x)

Rectangle sides
A = (y)(x)

Rectangle base
A = (8x)(y)


Area of tray = (1/2)(8x+10x)(x) + (1/2)(8x+10x)(x) + (y)(x) + (y)(x) + (8x)(y)


Simplify that down.
Now we have an expression in terms of A, x and y. However, if you look at what you are trying to reach, the expression is only in A and x.

You have to then substitute Part A into Part B.
So substitute whatever Y = blah blah for Y in part B.

Simplify and you should get the what you are aiming for.

So this is a really hard question that I'm stuck on, anyone care to help me. I can do part a but not the rest :/

HAAAA! what is that makes you think that I deserve such an award?

Out of interest, given that y = 900/9x^2 which divides down to 100/x^2, could this not be simplified through square rooting to 10/x ? hence y= 10/x

HAAAA! what is that makes you think that I deserve such an award?

Nothing at all bbygurl, noothing at all xoxo

shes YOLO
hahahahahahahahahhahahahahaahhhhhhhhhhhhaa

I might as well do the rest of the problem Cx

I just posted B.

Part C.
Stationary point = key word for "I have to differentiate something!"

A stationary points are the points where the derivative = 0. In this case, dA/dx =0.
It says you are looking for the value of x.

So you are looking for value of x when dA/dx = 0

So differentiate A= something something
Set it equal to 0
Solve for x.
REMEMBER 3 SIG FIGS

d.
The minimum value of A is easy to find.
You probably got only one answer for C.
Plug it back into A=blah blah and solve for A. Done


Now, to prove its a minimum point.
The best way to prove a minimum point is to take a second derivative.
So find out what d^2A/dx^2 is.

Take your X value and plug it into the second derivative.

This is the way I remember it!
If the second derivative is negative , the graph of the derivative it looks like a frowny face ^. That means its a maximum.
If the second derivate is positive, the graph of the derivative looks likea happy face U . The means its a minimum.

Anyway. You should get a positive answer.
So put
d^2A/dx^2 > 0 Therefore your x is a minimum ;P