Melody

"I multiplied by 30 instead of 68 and added 68 instead of 30.        

YES

"As for the latex, i can only divide the numerator and denominators and not numerator by numerator?"

YES

Please think about this example

$$\\\frac{3}{4}\times \frac{7}{8}\\
$You'd want to cancel the 4 with the 8 but you can't!! $\\
$you do know that it equals$\\
\frac{3*7}{4*8}=\frac{21}{32}\\\\
$21 and 32 have no common factors, this cannot be simplified!! $\\
$If this answer cannot be simplified then the working you used to get there cannot be simplified either!$$$

Do you want to try that question over again ?

Also, whenever you see fractions on the forum I would like you to race in and try to answer them.

You probably already do that as much as you can anyway :)

You just find the largest fraction there.   The only one that is bigger than 1/2 is 2/3 so that is the biggest.

All the others could be subsets of that 2/3  so the largest number that could not have any of these ingredients is 1- 2/3  =  1/3

1/3  of  24 cakes  is   8 cakes.  

I don't know what you did Titanium but I will give you credit for your effort :)

That IS funny Awesomeee     

Now I am going to show you have to reference a specific post withing a thread.  :)

At the top right of each post is a date stamp.  If you click on that the address in the address bar will change from being the address of the question at the top to the address of the specific post you clicked into.

Now you can just copy and paste the specific post address that you want.  Next time you want to reference the 15th post in your thread you will  be able to send me straight to it and I won't have to count anything.

Do you understand?    Use it a few times so you wont forget :)

Hi MG,

I am so pleased that you can use LaTex, it makes it so much easier to check your work. 

When you cancel fractions you can only do it with multiply (no mixed numerals either)  I think you already know that,

BUT when you cancel it has GOT to be something on the TOP with something on the BOTTOM

It cannot be 2 things on the top  OR two things on the bottom, it doesn't work that way.

That was a repetitive mistake that you made all the way through.  

It worries me a little that you cannot see the error on the second line too.  It does not effect the 3rd line but that is just a lucky conincidence.  Do you want to take a look again and see if you cannot spot the error :)

What do you want to do.   Do you want to leave it and do easier examples.

Maybe you could redo it but just add one or two lines at a time so I can check it as you go :/

Actually if you really can't see what is wrong with the second line then I should concentrat on that first.

It is important that you not only can get the answers and working correct but that you also completely understand why you are doing things and why they work.

--------------------

Hi Titanium,

Yes you have :))

@@ End of Day Wrap   Fri 5/6/15   Sydney, Australia Time   11:55am   ♪ ♫

Hi all,

Our fabulous answerers today were Radix, asinus, Jerico,Alan, MathsGod1 and CPhill.  Thank you  

Sat 6/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

 A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder that will require the least amount of material, that is the surface area, to make.

$$\\V=\pi r^2h=8\pi\\\\
r^2h=8\\\\
r^2=\frac{8}{h}\\\\
r=\sqrt{\frac{8}{h}}\\\\
$Let surface area=A$\\\\
A=2\pi rh+2\pi r^2\\\\
A=2\pi h\sqrt{\frac{8}{h}}+2\pi \frac{8}{h}\\\\
$This will be a max or a min when $\frac{dA}{dh}=0\\\\
A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\$$

$$\\A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\
A=4\sqrt2\pi h^{0.5}+16\pi h^{-1}\\\\
\frac{dA}{dh}=2\sqrt2\pi h^{-0.5}-16\pi h^{-2}\\\\
\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\\\$$

$$\\$Find stat points $\frac{dA}{dh}=0\\\\
2\sqrt2\pi h^{-0.5}-16\pi h^{-2}=0\\\\
\sqrt2 h^{-0.5}-8 h^{-2}=0\\\\
\sqrt2 h^{1.5}-8=0 \qquad $I mult both sides by $ h^2\\\\
h^{1.5}=\frac{8}{\sqrt2}\\\\
h=[\frac{8}{\sqrt2}]^{2/3}\\\\
h=\frac{2^2}{2^{1/3}}\\\\
h=2^{5/3}\\\\$$

$$\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\$$

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{1.5}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{3}}\right)} = {\mathtt{2.356\: \!194\: \!490\: \!192\: \!344\: \!8}}$$

GOOD    (I  was hoping  this answer would be positive, otherwise there would be an error)

$$If\; h=2^{5/3}\quad $A minimum amount of material will be used$$$

$$\\r=\sqrt{\frac{8}{h}}\\\\
r=\sqrt{\frac{2^3}{2^{5/3}}}\\\\
r=\sqrt{2^{4/3}}\\\\
r=2^{4/6}\\\\
r=2^{2/3}\\\\
r=\sqrt[3]{4}\\\\$$

So the minimum amount of material will be needed if      $$r=\sqrt[3]{4}\quad and \quad h=\sqrt[3]{32}$$

(The units are in feet)

This answer is identical to Alan's but our methods are a bit different     

MathsGod1, this is a good one for you :)