# A Match Box.... differentiation question

• lionely
In summary, the problem involves finding the area of material used to make a match box with an outer cover and a rectangular box inside. The volume of the box is given and the length is 1.5 times the breadth. The problem also asks to show that the least amount of material is used when the length of the box is approximately 3.7cm. The solution involves setting up an expression for the surface area of the box and differentiating it to find the critical points. The final answer is 2.46cm for the breadth and 3.7cm for the length.
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.

## 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)

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?

Last edited:
Take more care.

Differentiate.

Set equal to zero.

Solve.

Test each root to be max or min.

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

lionely said:

## 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.

## 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)

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.

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

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

Doing it my way brings mastery and independence.

Which do you want in life?

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

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.

lionely said:
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).

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)?

lionely said:
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.

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

lionely said:
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.

Have you found why you're not getting 3.7cm ?

( What is 2/3 of 3.7 ? )

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

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

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.

lionely said:
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.

Last edited:
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...

BvU said:
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.

ehild

## 1. What is a match box?

A match box is a small container, typically made of cardboard, that is used to store and carry matches for lighting fires. It usually has a striking surface on the side for igniting the matches.

## 2. How does a match box work?

A match box works by containing matches in a small, sealed space to keep them dry and protected. The striking surface on the side of the box creates friction when the match is swiped against it, causing the match head to ignite and light.

## 3. What are the different types of match boxes?

There are various types of match boxes, including cardboard match boxes, wooden match boxes, and plastic match boxes. There are also novelty match boxes that come in unique shapes and designs.

## 4. Are there any safety precautions when using a match box?

Yes, there are important safety precautions to take when using a match box. Always keep the box closed when not in use to prevent accidental fires. Only strike the match on the designated striking surface, and make sure to properly extinguish the match after use by running it under water or pressing it into a non-flammable surface.

## 5. How can a match box be environmentally friendly?

Match boxes can be environmentally friendly by using biodegradable materials such as paper or cardboard instead of plastic. Additionally, some match boxes are made with recycled materials. It is important to properly dispose of match boxes by recycling them or using them for other purposes, such as storing small items.

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