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https://a44694f152ee562e38a9454969d2b9a549a03543.googledrive.com/host/0B1ZiqBksUHNYOTR1Zk42aXdIazg/09%20Gold%201%20-%20C2%20Edexcel.pdf

help with question 7a because why not? :smile:

Edit: Actually i need help with part b

(edited 7 years ago)

Reply 1

IrrationalRoot

It should be clear that the total length of the 12 edges is

L=4(2x)+4x+4y=12x+4y

where y is the dimension of the cuboid that you aren't given.

But you know that the volume is 81, so

(x)(2x)(y)=81 \Rightarrow y=\dfrac{81}{2x^2}

. Therefore

L=12x+4y=12x+\dfrac{162}{x^2}

, as required.

EDIT: Not sure why part (b) should cause any problems. It's your typical differentiate, set equal to 0, solve, question.

(edited 7 years ago)

Reply 2

thefatone

Original post by IrrationalRoot

It should be clear that the total length of the 12 edges is

L=4(2x)+4x+4y=12x+4y

where y is the dimension of the cuboid that you aren't given.

But you know that the volume is 81, so

(x)(2x)(y)=81 \Rightarrow y=\dfrac{81}{2x^2}

. Therefore

L=12x+4y=12x+\dfrac{162}{x^2}

, as required.

Thanks for that what about part b then?

Reply 3

Zacken

Original post by thefatone

Thanks for that what about part b then?

You have an expression for

L

in terms of

x

, differentiate and set it equal to 0 to find the minimum points.

Reply 4

thefatone

Original post by Zacken

You have an expression for

L

in terms of

x

, differentiate and set it equal to 0 to find the minimum points.

derp, thanks so much, i'm forgetting everything, although things are coming back slowly to me ^-^

Reply 5

Zacken

Original post by thefatone

derp, thanks so much, i'm forgetting everything, although things are coming back slowly to me ^-^

No problem.

Reply 6

thefatone

Original post by Zacken

No problem.

where my answer was

x^3=27

then

x=\sqrt[3] 27[br][br]x=3\sqrt3

i don't need to put a

\pm

do i?

Reply 7

Zacken

Original post by thefatone

where my answer was

x^3=27

then

x=\sqrt[3] 27[br][br]x=3\sqrt3

i don't need to put a

\pm

do i?

No,

x^n

is injective for odd

n

. i.e: you only use the

\pm

when you have x^(even number)

Reply 8

thefatone

Original post by Zacken

No,

x^n

is injective for odd

n

. i.e: you only use the

\pm

when you have x^(even number)

i though so thanks :smile:

Reply 9

Math12345

Original post by thefatone

where my answer was

x^3=27

then

x=\sqrt[3] 27[br][br]x=3\sqrt3

i don't need to put a

\pm

do i?

Do you mean x=3?

Reply 10

Zacken

Original post by thefatone

i though so thanks :smile:

Awesome.

Reply 11

thefatone

Original post by Math12345

Do you mean x=3?

uh oh
i put in

\sqrt 27

into my calculator thanks :smile:

Original post by Zacken

Awesome.

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