d2jsp
Log InRegister
d2jsp Forums > Off-Topic > General Chat > Homework Help > Can Someone Help With This Math Please
Prev123Next
Add Reply New Topic New Poll
Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 10 2022 10:27pm
Quote (brigadier @ Dec 10 2022 09:35pm)
The most optional prism is a cube

The most optimal cylinder is one with h = 2r

I believe the most optimal shape would be a sphere, but it is too impractical to consider. Optimizing solely for the purpose of minimizing material is not practical. 1 cubic foot is sand weighs around 101lbs. Imagine 50 cubic feet of sand in a spherical container. It would never sit still and if it was ever moved down an incline then it could cause serious damage. —imagine parking on a steep hill without your emergency break on.

Cylinders and prisms are used so often because they are practical. Easy to stack easy to move easy to tesselate a space with minimal waste.

Volume cube : s^3

50 = s^3

50^(1/3) = s

3.68403149864 = s

Surface area cube = 6s^2

Surface area = 6(50^(1/3))^2 = 81.4325284978 sq ft

Volume cylinder = pi r^2 h

Note that h= 2r


Volume cylinder = pi r^2 (2r)

Volume cylinder = 2 pi r^3

50 = 2pi r^3

(50/2pi) = r^3

(50/2pi)^(1/3) = r

1.99647271233 = r

Surface area cylinder = 2 pi r^2 + 2pi r h

Surface area cylinder = 2 pi r^2 + 2pi r (2r)

Surface area cylinder = 6 pi r^2

Surface area cylinder = 6 pi (1.99647271233)^2

Surface area cylinder = 75.1325069828


:hail:
Member
Posts: 5,079
Joined: Mar 13 2021
Gold: 13.00
Dec 11 2022 03:42am
Quote (brigadier @ Dec 11 2022 03:35am)
The most optional prism is a cube


that is correct if we set the geometric theorem that the ratio volume:surface increases the more regular a body gets to be known. my question is though: can that be proven by geometry (without basic calculus or Lagrange optimization or similar means)? or do we just accept it, because it seems obvious?

EDIT: not that it matters for this task, just curious :D

This post was edited by PiXi31415 on Dec 11 2022 03:42am
Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 11 2022 10:02am
Quote (PiXi31415 @ Dec 11 2022 04:42am)
that is correct if we set the geometric theorem that the ratio volume:surface increases the more regular a body gets to be known. my question is though: can that be proven by geometry (without basic calculus or Lagrange optimization or similar means)? or do we just accept it, because it seems obvious?

EDIT: not that it matters for this task, just curious :D


i appreciate both efforts, i am in a basic geometry course though so went with the other answer :D will clarify in the future for what course this is for as to not waste anyone’s time
Member
Posts: 38,156
Joined: Feb 16 2009
Gold: 2,673.69
Dec 11 2022 11:10am
Quote (PiXi31415 @ Dec 11 2022 01:42am)
that is correct if we set the geometric theorem that the ratio volume:surface increases the more regular a body gets to be known. my question is though: can that be proven by geometry (without basic calculus or Lagrange optimization or similar means)? or do we just accept it, because it seems obvious?

EDIT: not that it matters for this task, just curious :D


Oof. There might be some roundabout proof, but none that I want to type on my phone nor that a student in this class would pay attention to long enough for it to make any sense.

Most instructors in these lower courses would use some type of self-discovery question to type to lead students to convince students that the fact is true.

You can show that a square is the most optimal rectangle (I would do this through pictures for lower classes)

This then would force us to think that the base of our prism should probably be a square. Then we have to figure out the height.

Choose a constant for a given surface area (anything positive should do).

Rewrite both equations for surface area and volume now that we know it should have a square base.

Solve the surface area formula for height.

Substitute that horrid expression into volume.

You should be a left with a cubic

Graph the cubic equation

The maximum should show the height must be equal to length of the side.


I am not sure I would ever do this in a lower class. Most students tune me out if I try to go over a proof. In my experience you end up discussing the problem with the one brilliant student while the other 29 zone out and feel dumb.

Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 16 2022 01:55pm
Quote (brigadier @ Dec 11 2022 12:10pm)
Oof. There might be some roundabout proof, but none that I want to type on my phone nor that a student in this class would pay attention to long enough for it to make any sense.

Most instructors in these lower courses would use some type of self-discovery question to type to lead students to convince students that the fact is true.

You can show that a square is the most optimal rectangle (I would do this through pictures for lower classes)

This then would force us to think that the base of our prism should probably be a square. Then we have to figure out the height.

Choose a constant for a given surface area (anything positive should do).

Rewrite both equations for surface area and volume now that we know it should have a square base.

Solve the surface area formula for height.

Substitute that horrid expression into volume.

You should be a left with a cubic

Graph the cubic equation

The maximum should show the height must be equal to length of the side.


I am not sure I would ever do this in a lower class. Most students tune me out if I try to go over a proof. In my experience you end up discussing the problem with the one brilliant student while the other 29 zone out and feel dumb.


4. A roof is shown in the diagram below. Calculate the length of the beam required to support the roof. Show all work. [2 marks]

https://imgur.com/a/rMCtwnG.png

This post was edited by RollOneUp on Dec 16 2022 01:58pm
Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 16 2022 01:59pm
Quote (RollOneUp @ Dec 16 2022 02:55pm)
4. A roof is shown in the diagram below. Calculate the length of the beam required to support the roof. Show all work. [2 marks]

https://imgur.com/a/rMCtwnG.png


sorry i couldnt figure out how to upload image other than this way, this is Trigonometry , I was able to get through the other questions but this one has confused me, please help :D
Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 16 2022 02:13pm
this is the answer i got , is this correct??

First, find the angle at the top

<A + <B + <C = 180

30 + 30 + <C = 180

180 - 60 = <C

<C = 120°

SinC = opp / hyp

Sin(120) = C / 4

CSin120 = 4

C = 4 / Sin120

C = 4.618802154

Rounded up to 4.62

C = 4.62 M

The length of the beam to support the roof will be 4.62 M
Member
Posts: 38,156
Joined: Feb 16 2009
Gold: 2,673.69
Dec 17 2022 12:40am
It looks like you used right angle trig

I can’t load the image on my phone so I have no idea.
Member
Posts: 858
Joined: Dec 23 2021
Gold: 508.69
Dec 17 2022 07:23am
Quote (brigadier @ Dec 17 2022 01:40am)
It looks like you used right angle trig

I can’t load the image on my phone so I have no idea.


they gave 2 sides, both 4m, and 2 angles both 30 degree , asked to find the side that wasn’t given , the top of the triangle is the angle that was missing , bottom of triangle is what i had to find , so i found the angle of the top first then used it to solve for the side

This post was edited by RollOneUp on Dec 17 2022 07:23am
Member
Posts: 5,079
Joined: Mar 13 2021
Gold: 13.00
Dec 17 2022 09:00am
Quote (RollOneUp @ Dec 16 2022 09:13pm)
^brigadier this is the answer i got , is this correct??

First, find the angle at the top

<A + <B + <C = 180

30 + 30 + <C = 180

180 - 60 = <C

<C = 120°

SinC = opp / hyp

Sin(120) = C / 4

CSin120 = 4

C = 4 / Sin120

C = 4.618802154

Rounded up to 4.62

C = 4.62 M

The length of the beam to support the roof will be 4.62 M


according to the picture:
the full triangle is not rectangular, so we have to divide it into rectangular triangles, if we want to use simple trigonometrics. this is easily achievable by just cutting it in the middle, like it is drawn in the picture (reason for that is that the triangle is symmetric). naming half (!) of the bottom side 'd', we can apply the definition of the cosine:
cos(30 deg) = d/4m, so d = 4m*cos(30 deg) = 3.4641 m
now we can get the full length of the beam by just doubling this number: 2d = 6.9282 m
Go Back To Homework Help Topic List
Prev123Next
Add Reply New Topic New Poll