A Match Box.... differentiation question

[lionely]

Homework Statement

a match box consists of an outer cover, open at both ends, into which slides a rectangular box without a top. The length of the box is one and a half times its breadth, the thickness of the material is negligible, and the volume of the box is 25cm^3 . If the breadth of the box is x cm, find , in terms of x, the area of the material used. Hence show that, if the least area of material is to be used to make the box , the length should be 3.7cm approximately.

Homework Equations

The Attempt at a Solution

So I have this so far:

25 = 3/2 x^2 h [Volume and h is the height]

For the surface area now, I figured I would find the surface area of the outer box(box1) and the inner box(box2)
and then add them together.

so I have this :

box1 = ((3x^2)/2 + xh) (2) = 3x^2 + 2xh

box 2 = (3x^2)/2 + xh(2) + (3xh/2)(2) = (3x^2)/2 + 5xh

box1 + box2 = (9x^2)/2 + 7xh [ h = (50)/(3x^2) , from the volume ]

Area(A) = (9x^2)/2 + 350/(3x)

dA/dx = 9x - (350)/(3x^2)

for max/min dA/dx = 0

9x = 350/(3x^2)

x^3 = 350/27
x = 2.45cm (which is incorrect)

Where did I go wrong?

 

[Dr. Courtney]

Take more care.

Define your variables explicitly.

Recognize your objective function.

Recognize your constraint function.

Use your constraint to reduce your objective function to one variable.

Differentiate.

Set equal to zero.

Solve.

Test each root to be max or min.

 

[lionely]

I basically did that, but is my reasoning about the areas correct?

 

[SammyS]

Homework Statement

a match box consists of an outer cover, open at both ends, into which slides a rectangular box without a top. The length of the box is one and a half times its breadth, the thickness of the material is negligible, and the volume of the box is 25cm^3 . If the breadth of the box is x cm, find , in terms of x, the area of the material used. Hence show that, if the least area of material is to be used to make the box , the length should be 3.7cm approximately.

Homework Equations

The Attempt at a Solution

So I have this so far:

25 = 3/2 x^2 h [Volume and h is the height]

For the surface area now, I figured I would find the surface area of the outer box(box1) and the inner box(box2)
and then add them together.

so I have this :

box1 = ((3x^2)/2 + xh) (2) = 3x^2 + 2xh

box 2 = (3x^2)/2 + xh(2) + (3xh/2)(2) = (3x^2)/2 + 5xh

box1 + box2 = (9x^2)/2 + 6xh [ h = (50)/(3x^2) , from the volume ]

Area(A) = (9x^2)/2 + 100/x

dA/dx = 9x - (100)/x^2

for max/min dA/dx = 0

9x = 100/x^2

x^3 = 100/9
x = 2.23cm (which is incorrect)

Where did I go wrong?

For one thing:
You added 2xh + 5xh and got 6xh.

Also, be careful to use parentheses where needed.

 

[Dr. Courtney]

Student: "Teacher, teacher, find my mistake."

Teacher: "Learn to take care, use a better problem solving method, and find your own mistake."

Doing it your way leads to dependence and less competence.

Doing it my way brings mastery and independence.

Which do you want in life?

 

[lionely]

Okay I will try again before I come back , thank you

 

[lionely]

Did it over about 3 times. I believe the surface area expression is correct. So I just don't understand why I can't get 3.7cm.

 

[SammyS]

Did it over about 3 times. I believe the surface area expression is correct. So I just don't understand why I can't get 3.7cm.

You also have a mistake in the area of the cover (box 1).

 

[lionely]

I don't see the mistake, aren't the areas of the top and bottom of the cover (3x/2)(x)(2) and the area of the sides are (x*h)(2)?

 

[SammyS]

I don't see the mistake, aren't the areas of the top and bottom of the cover (3x/2)(x)(2) and the area of the sides are (x*h)(2)?

No.

The side is the the longest dimension times the height.

 

[lionely]

So that means the expression for the area = (9x^2)/2 + (400)/(3x)

dA/dx = 9x - (400)/(3x^2)

for max/min dA/dx = 0

9x = 400/(3x^2)
x^3 = 400/27
x = 2.46cm .

I still can't get 3.7cm

 

[SammyS]

So that means the expression for the area = (9x^2)/2 + (400)/(3x)

dA/dx = 9x - (400)/(3x^2)

for max/min dA/dx = 0

9x = 400/(3x^2)
x^3 = 400/27
x = 2.46cm .

I still can't get 3.7cm

Look at the problem statement.

Added in Edit:
Have you found why you're not getting 3.7cm ?

( What is 2/3 of 3.7 ? )

 

[lionely]

2/3 of 3.7 is the answer I have... but why would my answer be 2/3 of the length?

 

[SammyS]

2/3 of 3.7 is the answer I have... but why would my answer be 2/3 of the length?

You're comparing your answer to the wrong dimension.

 

[lionely]

Ah I see now the problem I had was with the english. When I saw length I thinking about a piece of cardboard that long, but when they said length it meant... the length of the box. Thank you, also this might be quite stupid but I don't fully see how the area for the sides of the outerbox is the longest dimension times the height. Why isn't the breadth*height. I thought the outerbox didn't have ends so I couldn't do the former.

 

[SammyS]

Ah I see now the problem I had was with the english. When I saw length I thinking about a piece of cardboard that long, but when they said length it meant... the length of the box. Thank you, also this might be quite stupid but I don't fully see how the area for the sides of the outerbox is the longest dimension times the height. Why isn't the breadth*height. I thought the outerbox didn't have ends so I couldn't do the former.

It's just a matter of which sides are "missing". The two missing sides are those with smallest area. That's common for this type of matchbox.

 

[BvU]

Nice thread. "the problem I had was with the english" seems a bit strange to me, since your english appears excellent to me. To me it looks as if the step from what you read to what you imagined went awry. Would it have helped if you had made a sketch of the kind of box you had in mind ?

I like dr C's advice -- as a generic motto. Perhaps it should include "make a drawing".

Nowadays no one in his right mind smokes anymore and matchboxes are almost extinct. Goes to show problem authors should be more aware of the cultural context of their poor victims...

 

[Mark44]

Would it have helped if you had made a sketch of the kind of box you had in mind ?

I like dr C's advice -- as a generic motto. Perhaps it should include "make a drawing".

I couldn't agree more. On almost every applied calculus problem it's a good idea to make a sketch. Beginning students seem to be resistant to this idea. Granted, it takes more effort at first, but it's a false economy to save time by not drawing a sketch, and then waste much more time going down paths that lead to a wrong answer.

Also, I believe that having a sketch to look at utilizes the half of the brain that isn't used in symbolic calculations.