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# Person who understands math I choose you!!!!!

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Some more help requested for a weird question

According to USPS, a parcel can measure no more than 108 inches in length and girth combined. The length is designated to be the longest side and girth is the distance around a cross-section perpendicular to the length.

The questions that hold all the answers to life are...

1.Assuming the cross section is a square sketch the parcel???

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2. Write an equation to model the volume???

3. What is the maximum volume???

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4. What are the dimensions of the parcel that produces the max volume???

༼ =ε/̵͇̿̿/’̿’̿ ̿ ̿̿ ̿̿ ̿̿ ಥ_ಥ ༽

If possible could you show your steps and explain them because I would like to understand this as well and not just get the answer.

Thank you!!!

dom6547  Feb 19, 2018
edited by dom6547  Feb 19, 2018
edited by dom6547  Feb 20, 2018
edited by dom6547  Feb 20, 2018
edited by dom6547  Feb 20, 2018
Sort:

#1
+12143
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You have a picture.....use that as your sketch

2.volume = l x w x w = lw^2     (w= side of the square)

3. 54000 cu in   is max

4. l + 4w =180   or  l= 180-4w   substitute this in to #2

(180-4w)(w^2) = volume

(180w^2 - 4w^3)   a 'local' max will be where slope = 0

Derivative  360 w - 12w^2  = 0

w = 0 or 30    (throw out '0' obviously)

w=30  then l = 180-4w=60

volume max = l x w x w = 60 x 30 x 30 = 54000 cu in

ElectricPavlov  Feb 20, 2018
edited by Guest  Feb 20, 2018
#2
+68
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Sorry, I meant 108, not 180 my apologies. I edited it back to 108 and will try to solve it from what you gave me but I would definitely appreciate it if anyone answers the question again.

dom6547  Feb 20, 2018
#3
+68
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With 108 I got 11664 cu in^3

Cphill I noticed you began to try to solve the problem.  Did you get the same answer??? If not could you post your answer with your steps? If you get the same answer could you also post your steps???

Thank you so much!

dom6547  Feb 20, 2018
#4
+12143
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Then with the same methods   w = 18   l = 36   volume = 11664 cu in

ElectricPavlov  Feb 20, 2018
#7
+68
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Thank you!!!

dom6547  Feb 20, 2018
#5
+84389
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1. It looks like the picture pretty much shows the cross-section as being a square

2.  We have that  G + L   =  108  ⇒  G  =  108 - L

Assuming  that the  cross-section is a square....its side is 1/4 of the girth and can be expressed as:

G/4  =  (108 - L) / 4

So....the volume  can be expressed as   L [ (108 - L) / 4] ^2

So we have that

V  = L [ (108 - L) / 4]^2   =  L [ 27  - L/4]^2 = L [ L^2/16 - 27L/2  + 729 ]  =

L^3/16  - 27L^2//2  +  729L

To find the max volume, we can use some Calculus  or a graph

The graph seems easiest....here it is :

https://www.desmos.com/calculator/ybfbga97jl

3.  Looking at the graph, the maximum volume is [ again, assuming a square base ]  = 11664  in^3

4.  The Length, L  =  36 in

The Girth, G   =  108  - 36   =   72 in

So....the side of the square  is  1/4 of this =  18 in

So...the dimensions are   18 in x 18 in x 36 in   =  11664 in^3

Note that the restrictions have been met   Girth + Length  =  (4 * 18)  +  36   =   72 + 36  =  108 in

CPhill  Feb 20, 2018
#6
+68
+1

Thanks to you all so much!!!

dom6547  Feb 20, 2018

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