Find the range of possible dimensions for a volume

[Traced]

Homework Statement

The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436m^3

Relevant Equations

None

I understand how to get the dimensions that equal 8436m^3. What I don't understand is how to find the range of all possible dimensions.

I solved the inequality to get $6w^{3}-13w^{2}-5w-8436$

Using systematic guessing I found the root is x=4, so the factor is x-4.

Dividing (x-4) into $6w^{3}-13w^{2}-5w-8436$ gives me $6w^{2}+59w+703$

so, $6w^3-13w^{2}-5w-8436 = (w-12)(6w^{2}+59w+703)$.

this means that

w=12
l=3(12)+1=37
h=2(12)-5=19

v=lwh
v=8436

This is as far as I get. I don't understand how to find the range of all possible dimensions.

 

[jedishrfu]

It says the range of dimensions l,w,h of at least 8436 m^3

so l=1, w=1, h=1 gives 1 m^3 which is less than 8436 m^3

But I think you also have to consider the limits for l, w, h ie $l = 3w$ and $h = 2w - 5$

since h must be greater than 0 then 2w=5 meaning w>=xxx at a minimum and similarly for the maximums too.

 

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It says the range of dimensions l,w,h of at least 8436 m^3

so l=1, w=1, h=1 gives 1 m^3 which is less than 8436 m^3

But I think you also have to consider the limits for l, w, h ie $l = 3w$ and $h = 2w - 5$

since h must be greater than 0 then 2w=5 meaning w>=xxx at a minimum and similarly for the maximums too.

I'm still confused. Could you explain it a bit more please? Right now I have 3 dimensions, l=37, w=12 and h=19. The dimensions I have the minimum dimensions that will give a box of $8436m^{3}$. I'm confused by

But I think you also have to consider the limits for l, w, h ie $l = 3w$ and $h = 2w - 5$

I guess I could do something like this?

w>=12
l>=37
h>=19

I don't know if that is how the question is meant to be solved though.

 

[Mark44]

Homework Statement:: The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436m^3
Relevant Equations:: None

I understand how to get the dimensions that equal 8436m^3. What I don't understand is how to find the range of all possible dimensions.

I solved the inequality to get $6w^{3}-13w^{2}-5w-8436$

That's not an inequality -- it should include <, or >, or >=, or <=.
The actual inequality is $w(3w + 1)(2w - 5) \ge 8436$ ($m^3$).

What you solved was the equation $w(3w + 1)(2w - 5) = 8436$

Using systematic guessing I found the root is x=4, so the factor is x-4.

Dividing (x-4) into $6w^{3}-13w^{2}-5w-8436$ gives me $6w^{2}+59w+703$

so, $6w^3-13w^{2}-5w-8436 = (w-12)(6w^{2}+59w+703)$.

this means that

w=12
l=3(12)+1=37
h=2(12)-5=19

Assuming your algebra is correct, if w, l and h are as above, you'll get the minimum volume. If you increase w, that will cause l and h to increase, and result in a larger volume. So the range for w is $w \ge 12$. What would be the ranges for l and h?

v=lwh
v=8436

This is as far as I get. I don't understand how to find the range of all possible dimensions.

 

[Mark44]

I guess I could do something like this?

w>=12
l>=37
h>=19

I don't know if that is how the question is meant to be solved though.

I think you added this after I started my earlier reply. I believe you're on the right track here, but you still need to work in the constraints that l = 3w + 1, and h = 2w - 5.
So you'll have an inequality for w, but equations for l and h, both of which are in terms of w.

 

[haruspex]

found the root is x=4

You mean w-12, right?

how to find the range of all possible dimensions

Does the polynomial you found have any other real roots?
What does that tell you about the values it takes as you increase w?

 

[Traced]

I think you added this after I started my earlier reply. I believe you're on the right track here, but you still need to work in the constraints that l = 3w + 1, and h = 2w - 5.
So you'll have an inequality for w, but equations for l and h, both of which are in terms of w.

Sorry I'm still confused. I'm just not really understanding this.

 

[Mark44]

Sorry I'm still confused. I'm just not really understanding this.

From the problem statement in post 1:

The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width.

From post 3:

w>=12
l>=37
h>=19

The first inequality is correct, but l and h are constrained by the conditions in the problem statement. So w can be any value >= 12, but the values of l and h depend directly on the value of w.

 

[Traced]

From the problem statement in post 1:

From post 3:

The first inequality is correct, but l and h are constrained by the conditions in the problem statement. So w can be any value >= 12, but the values of l and h depend directly on the value of w.

Do you mean like this?
l>=3w+1 and h>=2w-5?
This way the length and the height have the constraints.

 

[Mark44]

Do you mean like this?
l>=3w+1 and h>=2w-5?
This way the length and the height have the constraints.

No. Here's what I mean:
$w \ge 12$
$l = 3w + 1$
$h = 2w - 5$
w can take on any value 12 or larger, but the two equations guarantee that l and h satisfy the given constraints. They wouldn't do so with the two inequalities you wrote.