C2 Differetiation

A wire is bent into the plane shape ABCDEA. Shape ABDE is a rectangle and BCD is a semicircle with diameter BD.
The area of the region enclosed by the wire is R m², AE = x metres, AB = ED = y metres. The total length of the wire is 2m.

a) Find an expression for y in terms of x.

This I've solved and the answer is y = 1 - x/2 - πx/4

b) Prove that R = x/8 (8-4x-πx)
This I have also done - FINALLY - thanks to zacken

Given that x can vary, using calculus and showing your working,

c) find the maximum value of R. (You do not have to prove that the value you obtain is a maximum)

For this I am stumped. I have got as far as

dR/dx = 1 - x - πx/4
let dR/dx = 0

so 1 - x - πx/4 = 0

x + πx/4 = 1

4x + πx = 4

x(4 + π) = 4

x = 4/(4 + π)
and then I have fed that back into the original equation for R

With R being:

R = x/8 (8-4x-πx)

Can someone just show me how the bit before the bracket is simplified too

How does x/8 become 1/(2)(4+pi)??

Help please

Reply 1

Zacken

Original post by christinajane

...

\displaystyle \frac{x}{8} = \frac{\frac{4}{4 + \pi}}{8} = \frac{4}{4 + \pi} \times \frac{1}{8} = \frac{1}{4 + \pi} \times \frac{1}{2} = \frac{1}{2(4 + \pi)}

Reply 2

JohnnyDavidson

Just by simplifying. I've got some key points explaining what's going on, in green on the right

Edit: zacken beat me to it haha

(edited 8 years ago)

image.jpg349.7KB

Reply 3

christinajane

Oh my god I was so close!

This whole question has had a lot of rearranging etc etc and has taken me a long time to get to grips with it and with a lot of help from you guys. Thank you thank you!

I hope these rules and tricks stay in my head and my confidence starts to build up so I can start to do them without asking for help!

Reply 4

christinajane

Zacken - I tried to give you rep but apparently I have to 'spread it around' first before I can!

(edited 8 years ago)

Reply 5

Zacken

Original post by christinajane

I'm sure they'll become second nature the more practice you do! :smile:

Original post by christinajane

Zacken - I tried to give you rep but apparently I have to 'spread it around' first before I can!

Don't worry about it. :lol:

Reply 6

christinajane

Original post by JohnnyDavidson

Just by simplifying. I've got some key points explaining what's going on, in green on the right

Edit: zacken beat me to it haha

Ok so next question - just something silly I need to clear up. I know these sound like stupid questions but suppose I just need to ask them in order to get my confidence up!

So I have got it to the point like you have in your picture.

R = 1/2(4 + pi) (8 - 16/4+ pi - 4pi/4+pi)

Now I know you need to get rid of the fractions by multiplying everything in the bracket by the fraction....

So this in the answers becomes
R = 1/2(4 + pi) (32 + 8pi - 16 - 4pi)/4+ pi

My question is - if I am multiplying every term in the bracket by 4+pi why is there still a 4+pi dividing everything in the bracket still, after multiplying out one of the two fractions, as the two fractions have the same denominator doesnt the fact that I am multiplying all the terms in the bracket mean that it cancels out the two fractions....Hope you can see what I am trying to get at....Here is the link for question 5 - mixed exercise...
https://0e90d90be05fe2865012661b731d0adbe5865fd8.googledrive.com/host/0B1ZiqBksUHNYT1JhVExXbEs2dk0/C2%20Chapter%2009.pdf

(edited 8 years ago)

Reply 7

Zacken

Original post by christinajane

...

Do you remember your common denominators?

\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}

?

This is what they are doing here, we have

8 - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi}

. We want the 8 to have the same denominator as the rest, so we rewrite it just as we have

\frac{1}{2} = \frac{2}{4}

.

Indeed, we get

\displaystyle 8 = \frac{8(4+\pi)}{(4 + \pi)}

. So we now get:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}8 - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi} = \frac{8(4+\pi)}{4 + \pi} - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi}\end{equation*}

Now that they are all in common denominators, the numerator becomes

Unparseable latex formula:

\displaystyle [br]\begin{equation*}8(4+\pi) - 16 - 4\pi = 32 + 8\pi - 16 - 4\pi = 16 + 4\pi = 4(4 + \pi)\end{equation*}

So the entire fraction becomes:

Unparseable latex formula:

\displaystyle\begin{equation*}8 - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi} = \frac{8(4+\pi)}{4 + \pi} - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi} = \frac{4(4+\pi)}{(4+\pi)} = 4\end{equation*}

Hence, what we have is that:

Unparseable latex formula:

\displaystyle\begin{equation*}R = \frac{1}{2(4+\pi)}\left(\frac{8(4+\pi)}{4 + \pi} - \frac{16}{4 + \pi} - \frac{4\pi}{4 + \pi}\right) = \frac{1}{2(4 +\pi)} \left(4\right) = \frac{2}{4 +\pi}\end{equation*}

Hope that makes sense, turns out typing with an injured hand takes time. :lol:

Reply 8

christinajane

Whoa yeah - common denominators I remember - I guess I just wouldnt have thought to apply them here!

They always seem to teach you - that if you have a fraction in an equation get rid of it by multiplying the relevant terms in the equation by its denominator of the fraction you're trying to get rid of, which is what I done, I thought quite rightly - but no!

Is there some sorta sign I should look out for if I see somehting like this - where to best multiply out and where to add in a common denominator and the multiply out...etc

I hope you don't think I am asking stupid questions - maths is soo tricky! Knowing when and when not to apply its rules - the order of which seem to be different depending on the problem - unless its just me and not really knowing what I am doing with the rules - which is probably more accurate!

Hope your hand is ok - what you do to it?

Do you work for NASA?

You sound like a genius! :smile:

(edited 8 years ago)

Reply 9

Zacken

Original post by christinajane

Is there some sorta sign I should look out for if I see somehting like this - where to best multiply out and where to add in a common denominator and the multiply out...etc

Hmm, not off the top of my head. It's just something that becomes second nature once you do enough practice.

I hope you don't think I am asking stupid questions - maths is soo tricky! Knowing when and when not to apply its rules - the order of which seem to be different depending on the problem - unless its just me and not really knowing what I am doing with the rules - which is probably more accurate!

I promise you that in a month or so, you'll be completely fluent. :biggrin:

Hope your hand is ok - what you do to it?

Haha, thanks! Nothing serious. :lol:

Do you work for NASA?

You sound like a genius! :smile:

Flattering, but I'm very far from one. :smile:

Reply 10

JohnnyDavidson

Original post by christinajane

Were you timesing through by (4 + pi ) in the numerators only? In which case it probably won't work, as you'd have to multiply EVERYTHING by (4 + pi), which would give you R (4 + pi ) on the LHS (basically you can never remove it).

The way Zacken has done it is the best way I can think of, as it this case it is only 8's denominator want to change, rather than multiply values. To complicate it even further there is also "rationalising the denominator" (unless this is what you meant?), but it's not the best one to use hear so I won't mention it for now :wink: