A teacher covered the exterior of a rectangular prism-shaped box that measured 8 inches by 9 inches by 10 inches

eddy2017

eddy2017

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A teacher covered the exterior of a rectangular prism-shaped box that measured 8 inches by 9 inches by 10 inches using one sheet of rectangular-shaped wrapping paper that measured 2 feet by 3 feet. There were no gaps or overlapping paper. How many square inches of wrapping paper were left over?

Choose an answer
  1. 144
  2. 380
  3. 484
  4. 622

1644705439065.png
I know this is about finding the are of a rectangular prism. the formula is

A = 2wl + 2lh + 2hw, where w is the width, the l is the length, and the h is the height.
I have:
[math]width =8 inches[/math][math]length= 9 inches[/math][math]height=10in[/math] First, I will find the area of the rectangular prism box
[math][B]A = 2(wl) + 2(lh) + 2(hw)[/math][math]A=2(8*9) + 2(9*10) + 2(10*8)[/math][math]A =[/math]2(72) +2(90) + 2(80) in^3[math][/math] A=114+180+160[math][/math]A=484 in^3$$
the area of the rectangular-prism box is 484 cubic inches.

any hint now?
thanks,
eddy[/B]
 
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I really don't think that knowing the volume would help to find the amount of wrapping paper needed.
let's see the perimeter .
[math]Perimeter = 2L + 2W[/math]
[math]= 2 × 8 + 2 ×9[/math]
[math]= 16 + 18[/math]
[math]= 34 inches[/math]perimeter=34 inches

I will need a hint to proceed.
 
Last edited:
Will I have to find the base of the triangular prism?. will it help in solving the problem?
 
eddy2017 said:
A teacher covered the exterior of a rectangular prism-shaped box that measured 8 inches by 9 inches by 10 inches using one sheet of rectangular-shaped wrapping paper that measured 2 feet by 3 feet. There were no gaps or overlapping paper. How many square inches of wrapping paper were left over?

Choose an answer
  1. 144
  2. 380
  3. 484
  4. 622

View attachment 31086
I know this is about finding the are of a rectangular prism. the formula is

A = 2wl + 2lh + 2hw, where w is the width, the l is the length, and the h is the height.
I have:
[math]width =8 inches[/math][math]length= 9 inches[/math][math]height=10in[/math] First, I will find the area of the rectangular prism box
[math][B]A = 2(wl) + 2(lh) + 2(hw)[/math][math]A=2(8*9) + 2(9*10) + 2(10*8)[/math][math]A =[/math]2(72) +2(90) + 2(80) in^3[math][/math] A=114+180+160[math][/math]A=484 in^3$$
the area of the rectangular-prism box is 484 cubic inches.

any hint now?
thanks,
eddy[/B]
Click to expand...
Eddy

You are forgetting your units again. Moreover, what is the FIND? What units do we need for the FIND?

Name your variables and keep track of your units.

[math]P = \text {area of paper in sq feet.}\\ A = \text {surface area of box in sq inches}\\ \ell = \text {length in inches}\\ w = \text {width in inches}\\ h = \text {height in inches}.[/math]
Can you calculate P?

You have a formula for the srface area of the box. Does it make sense in terms of units? Why?

[math]A = 2(\ell w + \ell h + hw).[/math]
Is there anything in that giving something with inches cubed? No. So volume is irrelevant.

Is there anything in that giving something with plain inches? No. So perimeter is irrelevant.

Some of the paper was used. What mathematical operation do you use to find how much is left?

Can you use the result you got about the surface area of the box directly? Why not?
 
Thank you Jeff. I will go to to work on your detailed info. Thank you so much. Your questions are an amazing lead.I will post it tomorrow though.
Have all my dear tutors a happy night!.
 
Okay, let's see
P stands here for area of paper
[math]Ap=l * w[/math][math]Ap=2ft * 3ft[/math][math]Ap= 6ft^2[/math]okay, the are of the paper equals 3ft^2.
 
you asked :
You have a formula for the surface area of the box. Does it make sense in terms of units? Why?
A=2(ℓw+ℓh+hw).
Well, Jeff, the units of the Surface Area are inches.
the surface area tells me how much has to be covered by the paper so it does makes sense to know the surface area.
but the paper area is in ft. I can convert ft to inches, though.
wrapping paper measures 2 feet by 3 feet.
We would convert the values in feet to inches
CONVERSION FACTOR
[math]1 ft = 12 inches[/math]
Length = 2 ft = 2 × 12 = 24 inches

Width = 3 ft = 3 × 12 = 36 inches

So, [math][B]area of wrapping paper[/B] = l *w = 24 in * 36 in =[B]864 square inches[/math].[/B]

I can find the surface area of the box. I have been given the units in inches. I have the area that the wrapping paper covers in inches too.
so it makes sense to find the the surface are of the box!
WIDTH= 8 IN
LENGTH=9IN
HEIGHT=10 IN
Now I can plug them in the formula
[math]SA=2(ℓw+ℓh+hw)[/math].
[math]SA= 2(9*8+9*10 +10*8)[/math]SA=2(72+90+80) [math][/math]SA=2(242)[math][B][/math]SA=484 in^2$$[/B]

so now i have the area of the wrapping paper and the surface area of the rectangular-prism box so, all i have to do is subtract
Area of wrapping paper - Surface area of the box.
864 square inches - 484 square inches
= 380 square inches.

Therefore, the amount of square inches of wrapping paper left over is 380 square inches.
I hope all this is good. I think we have left no stone unturned.
Jeff, I think i got it this time. your pointers were really good. they made me sit up and take notice.
are cubic inches mentioned?, no?, so what do you need volume for?
Is there anything in that giving something with plain inches? No. So perimeter is irrelevant.
I am printing these tips and pasting them on my desktop. Keep'em coming.!!!
They made me zero in on what to go for.
Thanks.
 
You know, Jeff, reflecting on your advice and I am gonna start crossing out the stuff that I think it is superfluous info in a word problem. based on what you have said, that will facilitate focus and will make for no distraction or wasted time over mumbo jumbo that really do not add to the solution. I will start practicing that. It takes practice to know what to keep and what to discard, but I will begin by focusing on things like the ones you mentioned: units.
Do these units offered something for the solution...,so an so forth.
 
Morning Eddy.

First things first. Congratulations on finding the correct answer.

I am trying to teach you general techniques for solving problems that involve unknown numbers.

(1) NAME YOUR RELEVANT VARIABLES AND SPECIFY UNITS.

Every tutor here has told you that. It organizes your thoughts.

Now I played a little trick on you. When I named the variables, I did not name the FIND. SK has told you repeatedly to identify what is to be found. I was hoping you might notice that. So lets name the variables the most useful way.

I also asked about units. You said the surface area is measured in inches. But you know that is wrong: area is measured in square inches or square meters or square feel. Paying careful attention units will help you avoid errors and wasted effort.

[math]R = \text {area of paper remaining in square inches; what is to be FOUND.}\\ P = \text {area of original paper in in square feet.}\\ A = \text {surface area of box in square inches.}\\ \ell = \text {length of box in inches} = 10.\\ h = \text {height of box in inches} = 9.\\ w = \text {width of box in inches} = 8.\\ L = \text {original length of paper in feet} = 3.\\ W = \text {original width of paper in feet} = 2. [/math]
What is the value of this exercise? First, it puts all the relevant information together in a concise way. Second, it tells you how many unknowns you have. Three. Thus, you will need three equations. Third, it warns you that you have more than one type of unit involved: two variables in square inches, one in square feet, three in inches, and two in feet. Thus, you will need to think carefully about units when setting up equations.

(2) SET UP EQUATIONS

Superficially, this seems easy once you look up the formula for the surface area of a rectangular prism. Won’t the following work

[math]P = LW = 3 * 2 = 6, \ R = P - A, \text { and }\\ A = 2( \ell h + \ell w + hw) = 2(10 * 9 + 10 * 8 + 9 * 8) = 2(90 + 80 + 72) = 2 * 242 = 484.[/math]
That seems to entail that R = 6 - 484 = - 478, which is nonsense. We did not think about units.

The first and third equations have units that make sense. Why?

The second equation is wrong. Why? How do you fix the second equation?

I do remember that you got the correct answer. Well done, but Dr. Peterson, Subhotosh Khan, Jomo, and I are trying to get you to use tools for systematic thinking routinely.

Identifying variables and unknowns, specifying units carefully, taking advantage of dimensional analysis, asking whether equations make sense in terms of units are all aids in systematic thinking, which avoids errors and wasted effort.
 
JeffM said:
Morning Eddy.

First things first. Congratulations on finding the correct answer.

I am trying to teach you general techniques for solving problems that involve unknown numbers.

(1) NAME YOUR RELEVANT VARIABLES AND SPECIFY UNITS.

Every tutor here has told you that. It organizes your thoughts.

Now I played a little trick on you. When I named the variables, I did not name the FIND. SK has told you repeatedly to identify what is to be found. I was hoping you might notice that. So lets name the variables the most useful way.

I also asked about units. You said the surface area is measured in inches. But you know that is wrong: area is measured in square inches or square meters or square feel. Paying careful attention units will help you avoid errors and wasted effort.

[math]R = \text {area of paper remaining in square inches; what is to be FOUND.}\\ P = \text {area of original paper in in square feet.}\\ A = \text {surface area of box in square inches.}\\ \ell = \text {length of box in inches} = 10.\\ h = \text {height of box in inches} = 9.\\ w = \text {width of box in inches} = 8.\\ L = \text {original length of paper in feet} = 3.\\ W = \text {original width of paper in feet} = 2. [/math]
What is the value of this exercise? First, it puts all the relevant information together in a concise way. Second, it tells you how many unknowns you have. Three. Thus, you will need three equations. Third, it warns you that you have more than one type of unit involved: two variables in square inches, one in square feet, three in inches, and two in feet. Thus, you will need to think carefully about units when setting up equations.

(2) SET UP EQUATIONS

Superficially, this seems easy once you look up the formula for the surface area of a rectangular prism. Won’t the following work

[math]P = LW = 3 * 2 = 6, \ R = P - A, \text { and }\\ A = 2( \ell h + \ell w + hw) = 2(10 * 9 + 10 * 8 + 9 * 8) = 2(90 + 80 + 72) = 2 * 242 = 484.[/math]
That seems to entail that R = 6 - 484 = - 478, which is nonsense. We did not think about units.

The first and third equations have units that make sense. Why?

The second equation is wrong. Why? How do you fix the second equation?

I do remember that you got the correct answer. Well done, but Dr. Peterson, Subhotosh Khan, Jomo, and I are trying to get you to use tools for systematic thinking routinely.

Identifying variables and unknowns, specifying units carefully, taking advantage of dimensional analysis, asking whether equations make sense in terms of units are all aids in systematic thinking, which avoids errors and wasted effort.
Click to expand...
Thanks for taking the time to advise me at length. I'm trying to hold fast to it. Practice makes near perfect so I will keep practicing and putting all the good advice given here by all.
 
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