another diff minimum prob :/ (1 Viewer)

[MrBrightside]

quick quiz: just a little test

simplify ( 12x+7) / 5x

5x(12x + 7) :S but i dont think thats it... I fail.

 

[MrBrightside]

its kinda a trick question , you could write it as ( 12x/ 5x) + 7/ (5x) = 12/5 + 7/ (5x)

but if you gave that in an hsc I would be willing to bet atleast 40% of people would incorrectly cancel the x and say ( 12 +7) / 5 = 19/5 { well maybe they wouldnt as you WOULD think that just a constant as the final answer would set off warning bells, but you get the idea } , it is not a WHOLE factor

rofl i was gona seperate em, but then i cbb typing it ahaha

 

[Absolutezero]

5x(12x + 7) :S but i dont think thats it... I fail.

You can't bring the 5x out the front, as you have no value on the other side of the equation to multiple. There is no equation. Even so, the 5x would only cancel out, not end up as a multiplier.

 

[MrBrightside]

2x*2x
=(2x)^2
=4x^2

Be careful!

wait wait so which ways which? i fikin hate algebra.

 

[MrBrightside]

You can't bring the 5x out the front, as you have no value on the other side of the equation to multiple. There is no equation. Even so, the 5x would only cancel out, not end up as a multiplier.

true

 

[hscmathstutor10]

wait wait so which ways which? i fikin hate algebra.

2x * 2x

we multiply the constants to get 2 * 2 = 4

we multiply the variables ( by index laws we add their powers ie x^(1+1) = x^2)

then we get 4x^2

 

[Absolutezero]

In order to get (2x)^2, think of it like this.

let 2x = T

T * T = T^2

resub: = (2x)^2 ------> because T was 2x, it must be squared as a whole, and not just the x

 

[MrBrightside]

ok so how would it = just 2x^2 ...I swear I read it on this site. it was under helpful 2 unit tips, it was a sticky thread.

btw the next part of the original qs is find the dimensions that will give the maximum volume :/ so lost. errr i hate max / min probs. >< I need more practice.

 

[hscmathstutor10]

how would it = just 2x^2 ...I swear I read it on this site. it was under helpful 2 unit tips, it was a sticky thread.

btw the next part of the original qs is find the dimensions that will give the maximum volume :/ so lost. errr i hate max / min probs. >< I need more practice.

2x * x would give 2x^2

for max/min problems we differentiate and set it equal to zero, this gives us the the turning points of the graph ( which is where, geometrically, the miniums and maxiums must occur ).

Its all just applying index laws and differentiation rules, this is why the basics are so important!!

so V= ( 125x -x^3) / 2

now the 2 on the bottom ( which gives a (1/2) factor ) is a constant, it doesnt depend on x, so when we are differentiating or integrating we can take a constant out the front, and differentiate/integrate what is left inside the bracket, then multiply the answer we get at the end by the constant.

so V= (1/2) (125x -x^3)
do differentiate boths sides with respect to x

d/dx (V) = (1/2) d/dx ( 125x -x^3) { and d/dx (V) is just dV/dx}

so dV/dx= (1/2) ( 125-3x^2)

and we set equal to zero for max/min

so (1/2) ( 125-3x^2) =0

we set equal factor equal to zero

(1/2) =0 -- makes no sense--> NO SOLUTION

(125-3x^2) =0
3x^2 = 125
x^2 = 125/3
x= +- sqrt ( 125/3 ) { BUT, x is a length, and must be posiitive to make any geometric sense , so take the + case}

x= + sqrt ( 125/3)

now we test, second derivative test is really easy to do

d^2 V / dx^2 = (1/2) d/dx ( 125-3x^2) = (1/2) ( -6x) = -3x

which is negative for all positive x, which means we have a concave down graph, and drawing a general picture means a maxium value

so x= sqrt (125/3) = sqrt (25 * 5 / 3) = 5 sqrt ( 5/3) cm is the value for x,

then going back to the constraint eqn on the surface area ie : 2x^2 + 4xh = 250

we can get h

rearranging
4xh= 250-2x^2
h= (250-2x^2) / 4x

= 250 - 2 ( 125/3) / 4 ( 5 sqrt(5/3) )
= 125/ [ 20 sqrt (5/3) ]
= 25/ [ 4 sqrt ( 5/3) ]

then we should always rationalise the answer

multiply top and bottom by sqrt (5/3)

we get h = 25 sqrt (5/3) / [4 * (5/3) ] = (15/4) sqrt (5/3) cm

 

[MrBrightside]

2x * x would give 2x^2

for max/min problems we differentiate and set it equal to zero, this gives us the the turning points of the graph ( which is where, geometrically, the miniums and maxiums must occur ).

Its all just applying index laws and differentiation rules, this is why the basics are so important!!

so V= ( 125x -x^3) / 2

now the 2 on the bottom ( which gives a (1/2) factor ) is a constant, it doesnt depend on x, so when we are differentiating or integrating we can take a constant out the front, and differentiate/integrate what is left inside the bracket, then multiply the answer we get at the end by the constant.

so V= (1/2) (125x -x^3)
do differentiate boths sides with respect to x

d/dx (V) = (1/2) d/dx ( 125x -x^3) { and d/dx (V) is just dV/dx}

so dV/dx= (1/2) ( 125-3x^2)

and we set equal to zero for max/min

so (1/2) ( 125-3x^2) =0

we set equal factor equal to zero

(1/2) =0 -- makes no sense--> NO SOLUTION

(125-3x^2) =0
3x^2 = 125
x^2 = 125/3
x= +- sqrt ( 125/3 ) { BUT, x is a length, and must be posiitive to make any geometric sense , so take the + case}

x= + sqrt ( 125/3)

now we test, second derivative test is really easy to do

d^2 V / dx^2 = (1/2) d/dx ( 125-3x^2) = (1/2) ( -6x) = -3x

which is negative for all positive x, which means we have a concave down graph, and drawing a general picture means a maxium value

so x= sqrt (125/3) = sqrt (25 * 5 / 3) = 5 sqrt ( 5/3) cm is the value for x,

then going back to the constraint eqn on the surface area ie : 2x^2 + 4xh = 250

we can get h

rearranging
4xh= 250-2x^2
h= (250-2x^2) / 4x

= 250 - 2 ( 125/3) / 4 ( 5 sqrt(5/3) )
= 125/ [ 20 sqrt (5/3) ]
= 25/ [ 4 sqrt ( 5/3) ]

then we should always rationalise the answer

multiply top and bottom by sqrt (5/3)

we get h = 25 sqrt (5/3) / [4 * (5/3) ] = (15/4) sqrt (5/3) cm

errmm, the answer saids 6.45 cm X 6.45 cm X 6.45 cm ..... I did get x = 6.45..from the first deriv, but now becasue its square base the other side of square is 6.45 as well, so now I have to work out h, by rearranging the formula.

 

[hscmathstutor10]

errmm, the answer saids 6.45 cm X 6.45 cm X 6.45 cm ..... I did get x = 6.45..from the first deriv, but now becasue its square base the other side of square is 6.45 as well, so now I have to work out h, by rearranging the formula.

ahh i know where I went wrong

250 - 2 ( 125/3) / 4 ( 5 sqrt(5/3) ) , this line,


the top bit = 500/3 , not 125 !!.

so h = ( 500/3) / [20 sqrt (5/3) ] = (25/ [ 3 sqrt (5/3)] ) = 25 sqrt ( 5/3) / [ 3 * 5/3] ( by rationalising ) = 5 sqrt ( 5/3 ) = 6.45


thats what you get for doing calculations in your head!

 

[MrBrightside]

2x * x would give 2x^2

so my original understanding of this was right. dam someone must of got it wrong on this site then.

 

[Absolutezero]

Could be. Nothing here is to be treated as absolute fact, and you should also be careful when using someone elses notes. Essentially:

2x * 2x = (2x)^2 ------> which subsequently can be expanded to 4x^2

OR

2x * x = 2x^2

It basically comes down to whether you have a bracket or not as to whether its an x or a 2x.

 

[MrBrightside]

hi maths tutor

I did part (a), and saw how many stupid mistakes i made lol

i am up to this part

V = 250x^2 - 2x^4 / 4x

how do you simplify that more (be precise in your explaining please) thanks.


[WDF i keep clicking the "Post reply" button instead of "post quick reply" button, losing all my text i just typed, so doing it twice now] (not the first time this has happened). this site is BADLY ergonomically / user - friendly designed, but i'm happy its working for now

 

[hscmathstutor10]

x * 2 ( 125 -x^2) / (2 *2 * x) // copied from a previous post //

FULLY factorise the top and bottom, and cancel the whole factors, there is not an easiesr why to explain it

and you mean -2x^3 im fairly sure

 

[MrBrightside]

so we have S= 2x^2 +4xh=250 ( this is correct ! )

and V= Ah = x^2h

now we have to rearrange the first for h,

so 4xh = 250 - 2x^2
h = (250-2x^2 ) / 4x


then subing into volume formula

V= x^2 [ (250-2x^2) /4x ]
= x(250-2x^2) / 4
= x * 2 ( 125 -x^2) / (2 *2)
= x (125-x^2) /2
= (125x -x^3 ) /2

you see the line in bold, wher ethe 2 * 2 are I put 2x(2) still correct?

and just to clarify. that first 2 in the denominator, it can be applied two times to the 250 and the 2x^2 (which is what I always thought)
however the x next to my 2 at in the denominator can only be used once to take out the first x^2 to make it x.

 

[hscmathstutor10]

you see the line in bold, wher ethe 2 * 2 are I put 2x(2) still correct?

and just to clarify. that first 2 in the denominator, it can be applied two times to the 250 and the 2x^2 (which is what I always thought)
however the x next to my 2 at in the denominator can only be used once to take out the first x^2 to make it x.

no that is not correct, the first "x" you see is on the numerator, not the denominator.

It would be so much easiest to explain it in person buts that limitations of the net

maybe I should have done [x * 2 ( 125 -x^2) ] / (2 *2) to emphasise that the x is on the top

 

[MrBrightside]

ahh man i think you misinterpreted my comment.

what i did was:

V = x^2 (250 - 2x^2) / 2x(2) (Because you said factorise top and bottom)

= x (125 - x^2) / 2 (canceled out the ^2 power of the top x and bottom x, the 2 makes the 250 125, and the 2 becomes a hidden 1 for the x)

= 125x - x^3 / 2 (expanded form)