IB maths

3a.
A cuboid has a rectangular base of width x cm and length 2x cm. the height of the cuboid is h cm. the total length of the edges of the cuboid is 72cm.
The volume, V, of the cuboid can be expressed as V= ax^2 - 6x^3.
Find the value of a. (3 marks)

hey guys, can i pls pls have some help?

so I got one mark for:
V= 2x^2 h = ax^2 - 6x^3
and one mark for:
12x + 4h = 72.

How do I find a, though???? How do I get that last mark???

(edited 4 years ago)

Reply 1

mikaelalrc

OP

8

3b.
Find the value of x that makes the volume a maximum.
Any help ASAP please?????

Reply 2

mikaelalrc

OP

8

@JackMoseley or someone else, are you here?

Original post by mikaelalrc

3a.
A cuboid has a rectangular base of width x cm and length 2x cm. the height of the cuboid is h cm. the total length of the edges of the cuboid is 72cm.
The volume, V, of the cuboid can be expressed as V= ax^2 - 6x^3.
Find the value of a. (3 marks)

hey guys, can i pls pls have some help?

so I got one mark for:
V= 2x^2 h = ax^2 - 6x^3
and one mark for:
12x + 4h = 72.

How do I find a, though???? How do I get that last mark???

Reply 3

JackMoseley

15

Original post by mikaelalrc

3a.
A cuboid has a rectangular base of width x cm and length 2x cm. the height of the cuboid is h cm. the total length of the edges of the cuboid is 72cm.
The volume, V, of the cuboid can be expressed as V= ax^2 - 6x^3.
Find the value of a. (3 marks)

hey guys, can i pls pls have some help?

so I got one mark for:
V= 2x^2 h = ax^2 - 6x^3
and one mark for:
12x + 4h = 72.

How do I find a, though???? How do I get that last mark???

Apologies for only just seeing this. My initial thought from there would be to rearrange to get h as the subject ( h=(72-12x)/4 ) and then sub this into your formula for V. This should then expand nicely just in terms of x, in which the coefficient of x^2 is your value for a.

Hope this makes sense?

Reply 4

JackMoseley

15

Original post by mikaelalrc

3b.
Find the value of x that makes the volume a maximum.
Any help ASAP please?????

A maximum occurs when the second derivative is less than 0.

You’ll need to differentiate V and find the stationary points (dy/dx = 0). After you find x, find the second derivative of V, and sub in your positive value of x (as it’s to do with lengths in this question) to check it is a maximum.

You might find these videos (more specifically the second one) helpful from TLMaths in this subject:
https://sites.google.com/site/tlmaths314/home/a-level-maths-2017/teaching-order/year-1/21-optimisation

Reply 5

mikaelalrc

OP

8

Thank you so much, but I still don't quite get it...

Following your advice, these are my workings:

h=(72-12x)/4
V = 2x^2 ((72 - 12x)/4)
576x^2 - 96 = ax^2 - 6x^3
576x^2 = ax^2 - 6x^3 - 96
576 = a - 6x + 96
480 = a - 6x
a = 480 + 6x

Im confuzzled??????

Original post by JackMoseley

Apologies for only just seeing this. My initial thought from there would be to rearrange to get h as the subject ( h=(72-12x)/4 ) and then sub this into your formula for V. This should then expand nicely just in terms of x, in which the coefficient of x^2 is your value for a.

Hope this makes sense?

Reply 6

mikaelalrc

OP

8

Thank you so much!!!!

Original post by jackmoseley

a maximum occurs when the second derivative is less than 0.

You’ll need to differentiate v and find the stationary points (dy/dx = 0). After you find x, find the second derivative of v, and sub in your positive value of x (as it’s to do with lengths in this question) to check it is a maximum.

You might find these videos (more specifically the second one) helpful from tlmaths in this subject:
https://sites.google.com/site/tlmaths314/home/a-level-maths-2017/teaching-order/year-1/21-optimisation

Reply 7

JackMoseley

15

Original post by mikaelalrc

Thank you so much, but I still don't quite get it...

Following your advice, these are my workings:

h=(72-12x)/4
V = 2x^2 ((72 - 12x)/4)
576x^2 - 96 = ax^2 - 6x^3
576x^2 = ax^2 - 6x^3 - 96
576 = a - 6x + 96
480 = a - 6x
a = 480 + 6x

Im confuzzled??????