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Maths Question Help - GCSE!!!

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Old 22-04-2009, 07:25 PM
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RS-Leigh
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Default Maths Question Help - GCSE!!!

Need some help please guys, the wife is helping our niece revise for her gcse but we are struggling to remember these equations......

q1. for the equation, y=5000x - 625 xsquared, find dy/dx?

q2. find the co-ordinates of the turning point on the graph

y=5000x - 625 xsquared

Thanks

Leigh
Old 22-04-2009, 07:27 PM
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Christian and Beccy
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Fuck knows.
Old 22-04-2009, 07:27 PM
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Holy macaroni batman, WTF.

I think maybe you're over-estimating the intelligence levels on here.
Old 22-04-2009, 07:28 PM
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RS-Leigh
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Originally Posted by Christian and Beccy
Fuck knows.

Thanks

I just told my neice " we have a reply, dont panic"

Regards

leigh
Old 22-04-2009, 07:32 PM
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Originally Posted by Christian and Beccy
Fuck knows.
+ 1 !!!!
Old 22-04-2009, 07:38 PM
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Come on Leigh!


THINK its easy!














Think again its not
Old 22-04-2009, 07:47 PM
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jeez, this takes me back, its about differentiation between expressions blah blah blah, give me a few mins to try and remember and i will get back to you

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Old 22-04-2009, 07:47 PM
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Originally Posted by Luca
Come on Leigh!


THINK its easy!














Think again its not
Come on Luke, you must remember this.......its been 20 years since I done 'o' level maths

There is a second question coming soon......

Leigh
Old 22-04-2009, 07:48 PM
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Originally Posted by Twellsie
jeez, this takes me back, its about differentiation between expressions blah blah blah, give me a few mins to try and remember and i will get back to you
Thanks mate, this question alone is worth nearly 5% of the whole gcse exam

Leigh
Old 22-04-2009, 07:50 PM
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Originally Posted by RS-Leigh
Need some help please guys, the wife is helping our niece revise for her gcse but we are struggling to remember these equations......

q1. for the equation, y=5000x - 625 xsquared, find dy/dx?

q2. find the co-ordinates of the turning point on the graph

y=5000x - 625 xsquared

Thanks

Leigh
well what you do is y5000x multiple x squared divided by 5000yx = + -dy half it by dx= fuck noes
Old 22-04-2009, 07:52 PM
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have you got a graphical calculator?
Old 22-04-2009, 07:53 PM
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Originally Posted by Twellsie
have you got a graphical calculator?
no mate, just scentific

Leigh
Old 22-04-2009, 07:56 PM
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Wheres that Jimbo ???

Lets see if he fell asleep in this lesson
Old 22-04-2009, 07:56 PM
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Originally Posted by RS-Leigh
no mate, just scentific

Leigh
lol, i am going to have to remember how to do it properly then, i think i am getting there anyway, lol

p.s why the hell do kids need to know this?? lol
Old 22-04-2009, 07:56 PM
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y = 5000x - 625x^2
First find dy/dx = 5000-1250x

Now there are turning points when dy/dx = 0, so:

0 = 5000 - 1250x
1250x = 5000
x = 500/125 or 4

To find the y-coordinate of the turning point, put this value of x into the original function:

y = 5000x - 625x^2
= 5000(500/125) - 625 (500/125)^2
=20000 - 10000
= 10000

So there is a turning point at (4, 10000)

I'm pretty sure this is right.
Rich
Old 22-04-2009, 08:04 PM
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How do you jump to this..

First find dy/dx = 5000-1250x

????

Thanks

leigh
Old 22-04-2009, 08:05 PM
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Originally Posted by Sonic Boom
y = 5000x - 625x^2
First find dy/dx = 5000-1250x

Now there are turning points when dy/dx = 0, so:

0 = 5000 - 1250x
1250x = 5000
x = 500/125 or 4

To find the y-coordinate of the turning point, put this value of x into the original function:

y = 5000x - 625x^2
= 5000(500/125) - 625 (500/125)^2
=20000 - 10000
= 10000

So there is a turning point at (4, 10000)

I'm pretty sure this is right.
Rich
i am pretty sure you are right, that is the answer i got on my graphical calculator, i was just struggling, to work it back, lol
Old 22-04-2009, 08:09 PM
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We are struggling with this part

____________

y = 5000x - 625x^2
First find dy/dx = 5000-1250x

Now there are turning points when dy/dx = 0, so:

0 = 5000 - 1250x
1250x = 5000
x = 500/125 or 4

_____________

We see that you have divided both sides by x, but should it not then be

5000-625x ??


Thanks

leigh
Old 22-04-2009, 08:13 PM
  #19  
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Originally Posted by Luca
Wheres that Jimbo ???

Lets see if he fell asleep in this lesson
cheeky bastard

i just done my gcse's, well 10months ago, and got an A in maths

but this !! is way past me
Old 22-04-2009, 08:17 PM
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Originally Posted by J1mbo
cheeky bastard

i just done my gcse's, well 10months ago, and got an A in maths

but this !! is way past me
Come on Jimbo, this is a past paper from 2006.....

Leigh
Old 22-04-2009, 08:19 PM
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Y = 5000x – 625xsquared

First dy/dx = 0
So divide both side by x

y/x = (5000x / x) – (625x2/x) = 0

cancel the x’s out
0 = 5000 – 625x

Divide all parts by 625
0/625 = (5000/625) – (625x/625)

simplify
0 = 8 – x

Plus x to both sides

X = 8

No substitute x = 8 into original equation

Y = (5000 * 8) – (625 * 8)

Y = 40000 – 5000

Y = 35000

hope this helps
Old 22-04-2009, 08:21 PM
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sry forgot to square 8 on last part of equation, here it is again

Y = 5000x – 625xsquared

First dy/dx = 0
So divide both side by x

y/x = (5000x / x) – (625x2/x) = 0

cancel the x’s out
0 = 5000 – 625x

Divide all parts by 625
0/625 = (5000/625) – (625x/625)

simplify
0 = 8 – x

Plus x to both sides

X = 8

No substitute x = 8 into original equation

Y = (5000 * 8) – (625 * 8squared)

Y = 40000 – 40000

Y = 0
Old 22-04-2009, 08:23 PM
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Originally Posted by Drakey
Y = 5000x – 625xsquared

First dy/dx = 0
So divide both side by x

y/x = (5000x / x) – (625x2/x) = 0

cancel the x’s out
0 = 5000 – 625x

Divide all parts by 625
0/625 = (5000/625) – (625x/625)

simplify
0 = 8 – x

Plus x to both sides

X = 8

No substitute x = 8 into original equation

Y = (5000 * 8) – (625 * 8)

Y = 40000 – 5000

Y = 35000

hope this helps
Beat me too it.
Old 22-04-2009, 08:25 PM
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Oh Leigh


Jimbo asked whens your neices Prom
Old 22-04-2009, 08:25 PM
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We have sussed it out using half your answer and half sonic booms answers

No substitute x = 8 into original equation

Y = (5000 * 8) – (625 * 8)

Y = 40000 – 5000

Y = 35000

hope this helps

You forgot to square the 625 8x ^2

so

40000 - 625 x 64 =y

y = 40000 - 40000 =0

So are we correct in saying

x = 8

y = 0
Old 22-04-2009, 08:27 PM
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i did repost it correctly just after my last post but not quick enough lol
Old 22-04-2009, 08:27 PM
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Part 2

Is the turning point a max or a min?

Gut feeling its a min due to the negative value in the equation

Thanks for the help so far

Leigh
Old 22-04-2009, 08:28 PM
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Originally Posted by Drakey
i did repost it correctly just after my last post but not quick enough lol
Just noticed mate, apologies, I have the wife and neice talking in my ear and I cant multi task!!!!
Old 22-04-2009, 08:31 PM
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Originally Posted by Luca
Oh Leigh


Jimbo asked whens your neices Prom
Thanks for your input Luke, you are clearly as good at maths as you are at launching your car of the line

Leigh

Last edited by RS-Leigh; 22-04-2009 at 08:49 PM.
Old 22-04-2009, 08:31 PM
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if Y is a positive value then its the minimum turning point, i think y = 0 is classed as a positive value but not 100%
Old 22-04-2009, 08:32 PM
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What did I manage to balls up?
Rich
Old 22-04-2009, 08:33 PM
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Originally Posted by Drakey
if Y is a positive value then its the minimum turning point, i think y = 0 is classed as a positive value but not 100%
Yes 0 is a positive number
Rich
Old 22-04-2009, 08:35 PM
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many thanks for the help guys

She is on her way home but is back tomorrow night as we have helped her so much

Stand by for the next installment tomorrow

Thanks

Leigh
Old 22-04-2009, 08:39 PM
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Oh I seen my error I actually times it by 2. What a numpty!!!
Rich
Old 22-04-2009, 08:42 PM
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Originally Posted by RS-Leigh
We are struggling with this part

____________

y = 5000x - 625x^2
First find dy/dx = 5000-1250x

Now there are turning points when dy/dx = 0, so:

0 = 5000 - 1250x
1250x = 5000
x = 500/125 or 4

_____________

We see that you have divided both sides by x, but should it not then be

5000-625x ??


Thanks

leigh
there is no need, you have x on both side of the equation, so that leaves you with

5000 - 625^2, ie 5000 - 1250
Old 22-04-2009, 08:51 PM
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Originally Posted by Twellsie
there is no need, you have x on both side of the equation, so that leaves you with

5000 - 625^2, ie 5000 - 1250
but its 5000x - 625 x^2 (x squared)

so if you divide both sides by x then its

5000 - 625x

I think

Leigh
Old 22-04-2009, 08:52 PM
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lol, i am doing the same thing, x2 instead of to the power of 2
Old 22-04-2009, 08:55 PM
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Originally Posted by RS-Leigh
but its 5000x - 625 x^2 (x squared)

so if you divide both sides by x then its

5000 - 625x

I think

Leigh
shit no, you are right, its the x that is squared
Old 22-04-2009, 08:56 PM
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I used to be really clever, Grade B in maths I got... But let me ask, WHAT THE HELL ARE YOU ALL ON ABOUT?
Old 22-04-2009, 08:57 PM
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Christ, that totally re-affirms why I hate maths. What possible use is that stuff in the real world? Is there any time when you'd actually need to do something like that? (Serious question!)


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