# Thread: Basic Application of Differentiation Question

1. ## Basic Application of Differentiation Question

I think it's basic I'm studying it at high school and I'm led to believe that this shouldn't be that difficult!

The question reads: A rectangular box is to be made to the following requirements:
1) The length must be one and a half times the width. 2) The twelve edges must have a total length of 6m.
Find the dimensions of the box that meets these requirements and that maximize the volume.

I have the answer at the back of the book if that'd help out anyone? Thanks for your help I really appreciate it.

My (failed) attempts so far: (Math's isnt my forte)
1.5(4w) + 4w +4h = 6m
then got it down to 1w = (6-4h)/10

Subbed it into the formula I made: 1.5((6-4h)/10) + 4((6-4h)/10) +4h = 6 but it ended up saying x=x on my calc. I am lost!

2. Originally Posted by camjsiv
I think it's basic I'm studying it at high school and I'm led to believe that this shouldn't be that difficult!

The question reads: A rectangular box is to be made to the following requirements:
1) The length must be one and a half times the width. 2) The twelve edges must have a total length of 6m.
Find the dimensions of the box that meets these requirements and that maximize the volume.

I have the answer at the back of the book if that'd help out anyone? Thanks for your help I really appreciate it.

My (failed) attempts so far: (Math's isnt my forte)
1.5(4w) + 4w +4h = 6m
then got it down to 1w = (6-4h)/10

Subbed it into the formula I made: 1.5((6-4h)/10) + 4((6-4h)/10) +4h = 6 but it ended up saying x=x on my calc. I am lost!
$V=w.h.l=1.5w^2h$

The edge length constraint is:

$4w+4l+4h=4w+6w+4h=10w+4h=6$

So substituting $h=(6-10w)/4$ into the volume gives:

$V=1.5w^2(6-10w)/4$

To find the width corresponding to maximum volume solve:

$\frac{d}{dw}V=0$

for $w$ and then choose the solution corresponding to maximum volume and obtain the other dimensions from that.

CB