1. ## my final project

if anyone could help asap that would be nice

i have to find out the sizes and volume and stuff then actualy build it.

i have to make a rectangular pism with a surface area of 9 in. can you find the base side lengths and height?

then i have to make a cylindrical package that holds 150 gummy bears when each gummy bear is .24 in cubed. i need again side lengths and what i need to build it.

last one i need a regular hexagonal pyramid with a volume of 196 cm squared. and i need the base size and side lengths.

if you could show me how the math is, like how to work it out, so i can know how to do this stuff...

this is realy importand i will fail my class if i can't figure this out. please help me...

ok well i guess this is realy hard or something so here are the formulas and stuff.

Prism
Lateral Area: perimiter(height)
Surface Area: 2(base) + Lateral Area
Volume: Base(Height)

cylinder
Lateral Area: circomference(height)
Surface Area: 2(base) + lateral area
Volume: base(height)

pyramids
Lateral Area: 1/2 perimeter (slant height)
Surface Area: Base + lateral area
Volume: 1/3 base (height)

area of a regular polygon: 1/2 apothem (peremeter)

2. Originally Posted by tracyneedshelp
if anyone could help asap that would be nice

i have to find out the sizes and volume and stuff then actualy build it.

i have to make a rectangular pism with a surface area of 9 in. can you find the base side lengths and height?
there are many rectangular prisms we can build to fit this criteria. you have to be more specific as to how you want the measurements of the side to relate to each other. if, for instance, you could give us any two measurements, we could find the third. say you give the length and width (within reason), we can tell you what the height has to be to have a surface area of 9 in^2 , etc

then i have to make a cylindrical package that holds 150 gummy bears when each gummy bear is .24 in cubed. i need again side lengths and what i need to build it.
this means we have to make a cylindrical package with a volume of at least 36 in^3 (i say at least because we may have to account for the extra spaces between the gummy bears when they are put into the package. you will need two circular disks and a rectangle sheet to build this container. again, more info is needed before we give you exact measurements.

last one i need a regular hexagonal pyramid with a volume of 196 cm squared. and i need the base size and side lengths.
again, give us something to work with. tell us the height that you want the pyramid to be, and we will show you how to find the length of all the sides or vice versa.

if you could show me how the math is, like how to work it out, so i can know how to do this stuff...
you already have the formulas, now you need to figure out the measurements. plug in the values in the formula that you must have. then plug in all the values except one as you would like them to be, then solve for the last one.

Formulas:

Prism
Lateral Area: perimiter(height)
Surface Area: 2(base) + Lateral Area
Volume: Base(Height)

cylinder
Lateral Area: circomference(height)
Surface Area: 2(base) + lateral area
Volume: base(height)

pyramids
Lateral Area: 1/2 perimeter (slant height)
Surface Area: Base + lateral area
Volume: 1/3 base (height)

area of a regular polygon: 1/2 apothem (peremeter)

3. i know there are many different ways that's the point to just find one way...i do know how to find one thing if given 2 pieces of info, but i don't know how to take one and make the other two...you see my problem?

that's all the info the teacher gave me...

4. Originally Posted by tracyneedshelp
i know there are many different ways that's the point to just find one way...i do know how to find one thing if given 2 pieces of info, but i don't know how to take one and make the other two...you see my problem?

that's all the info the teacher gave me...
my concern is, do you want to build all of these objects to fit together somehow, or should they be separate? if so, we can strat working on finding any one way to do each

5. no they're just seperate...

see i have the formulas, but i don't know how to use them with just one # to plug in.

is there a way to figure this out besides tiral and error or no?

6. Originally Posted by tracyneedshelp
no they're just seperate...

see i have the formulas, but i don't know how to use them with just one # to plug in.

is there a way to figure this out besides tiral and error or no?
fine, let's tackle the first:

i have to make a rectangular pism with a surface area of 9 in. can you find the base side lengths and height?
we are given the formulas:

Prism
Lateral Area: perimiter(height)
Surface Area: 2(base) + Lateral Area
Volume: Base(Height)

we want the surface area to be 9 in^2

=> 9 = 2(base) + Lateral Area

if we let the area of the base be 3 and the lateral area be 3, we have the desired result, since 9 = 2(3) + 3

so Area of base = 3
and Lateral area = 3

now let's just choose the length and width of the base so the area of the base becomes 3

we can choose length = 2in and width = 1.5 in

so Area of base = length*width = 2(1.5) = 3

now the lateral area = perimter*height

since the length = 2 and the width = 1.5, the perimeter is 7

=> lateral area = 3 = 7*height
=> height = 3/7 in

so those are the measurements we can use for our prism.

length = 2
width = 1.5
height = 3/7

do you understand? try the others

7. uhm ok lets see with the cylinder it should be

.24 (150) to get the total volume right?

which is 36...

so... 36= Bh

uhm...but for this part i just have to do trial and error?

would it work if we say r= 3 so the C= 9pie= 28.27

so base = 28.27

ok i'm lost again...

8. Originally Posted by tracyneedshelp
uhm ok lets see with the cylinder it should be

.24 (150) to get the total volume right?

which is 36...

so... 36= Bh

uhm...but for this part i just have to do trial and error?

would it work if we say r= 3 so the C= 9pie= 28.27

so base = 28.27

ok i'm lost again...
No, start from the bigger quantities and work your way in.

let the base be 6 and the height be 6, now work out the radius

or you could let the base be 3 and the height be 12 and then work out the radius

9. B=6
H=6

6= pie r ^2

1.9= r^2

1.3=C

ok right so far i hope...

V= 1.3 pie r^2 (6)
V= 6 (6)
V=36

was it realy that easy? or did i mess up?

the next one i know will take me a while to get...i hate working with apothems...

10. Originally Posted by tracyneedshelp
B=6
H=6

6= pie r ^2

1.9= r^2

1.3=C

ok right so far i hope...

V= 1.3 pie r^2 (6)
V= 6 (6)
V=36

was it realy that easy? or did i mess up?

the next one i know will take me a while to get...i hate working with apothems...
that equation for the volume is incorrect. and i would not use decimals, keep things exact so you get the exact answer.

$\displaystyle B = 6$
$\displaystyle H = 6$

$\displaystyle \Rightarrow B = 6 = \pi r^2$

$\displaystyle \Rightarrow r^2 = \frac {6}{ \pi}$

$\displaystyle \Rightarrow r = \sqrt { \frac {6}{ \pi}}$

Remeber, to build a cylinder, we need two circular disks and a rectangular sheet to form the sides of the cylinder, the demensions of the rectangular sheet will be: length = height of the cylinder and width = circumference of the circular base.

since $\displaystyle r = \sqrt { \frac {6}{ \pi}}$

The circumference $\displaystyle C$ will be given by:

$\displaystyle C = 2 \pi r = 2 \pi \sqrt { \frac {6}{ \pi}}$

And you can simplify that if you want.

So finally, to make the required cylinder we need a rectangular sheet with height = 6 and width = $\displaystyle 2 \pi \sqrt { \frac {6}{ \pi}}$

and we also need two circular disks with each of their radii = $\displaystyle \sqrt { \frac {6}{ \pi}}$

11. [size=3]Hello, Tracy!

i have to make a rectangular prism with a surface area of 9 inē.
can you find the base side lengths and height?

As Jhevon pointed out, there is no one answer to this problem.

A rectangular prism is simply a "box" with length, width and height.
Code:
         *- - - - - - - -*
/               /|
/               / |H
* - - - - - - - *  |
|               |  *
H|               | /
|               |/ W
* - - - - - - - *
L

The front and back panels have area $\displaystyle LH$ each: .$\displaystyle 2LH$

The top and bottom panels have area $\displaystyle LW$ each: .$\displaystyle 2LW$

The left and right panels have area $\displaystyle WH$ each: .$\displaystyle 2WH$

The total surface area is: .$\displaystyle 2LH + 2LW + 2WH \:=\:9$

We can use any three values of $\displaystyle L,\,W,\,H$ that satisfy the equation.

If we solve for $\displaystyle L$, we get: .$\displaystyle L \:=\:\frac{9 - 2WH}{2(W + H)}$

Let $\displaystyle W = 1,\,H = 1$, then: .$\displaystyle L = \frac{7}{4}$
. . A box with dimensions: $\displaystyle 1 \times 1 \times \frac{7}{4}$ will have a surface area of 9 inē.

Let $\displaystyle W = 1,\,H = 2$, then: .$\displaystyle L = \frac{5}{6}$
. . A box with dimensions: $\displaystyle 1 \times 2 \times \frac{5}{6}$ will have a surface area of 9 inē.

As you can see, the possibilities are endless.

12. Originally Posted by Soroban
[size=3]Hello, Tracy!

As Jhevon pointed out, there is no one answer to this problem.

A rectangular prism is simply a "box" with length, width and height.
Code:
         *- - - - - - - -*
/               /|
/               / |H
* - - - - - - - *  |
|               |  *
H|               | /
|               |/ W
* - - - - - - - *
L

The front and back panels have area $\displaystyle LH$ each: .$\displaystyle 2LH$

The top and bottom panels have area $\displaystyle LW$ each: .$\displaystyle 2LW$

The left and right panels have area $\displaystyle WH$ each: .$\displaystyle 2WH$

The total surface area is: .$\displaystyle 2LH + 2LW + 2WH \:=\:9$

We can use any three values of $\displaystyle L,\,W,\,H$ that satisfy the equation.

If we solve for $\displaystyle L$, we get: .$\displaystyle L \:=\:\frac{9 - 2WH}{2(W + H)}$

Let $\displaystyle W = 1,\,H = 1$, then: .$\displaystyle L = \frac{7}{4}$
. . A box with dimensions: $\displaystyle 1 \times 1 \times \frac{7}{4}$ will have a surface area of 9 inē.

Let $\displaystyle W = 1,\,H = 2$, then: .$\displaystyle L = \frac{5}{6}$
. . A box with dimensions: $\displaystyle 1 \times 2 \times \frac{5}{6}$ will have a surface area of 9 inē.

As you can see, the possibilities are endless.

Thanks for that systematic approach Soroban, for some reason I was in trial and error mode

13. so the width would be 8.68 which rounds to 8.7

but what is sqrt?

...and the circumference of the two end is still 6 right?

for the last one...will it work if i have the radius=5 a=6 and each side be 8? with a height of 8.8 and it be right for the volume?

14. ok i'm starting to think i'm just stupid...cuz you guys are giving me all this help and i still have no idea how to d this stuff...

sorry guys for being so stupid :'[

15. Originally Posted by tracyneedshelp
so the width would be 8.68 which rounds to 8.7
yes, but as i said, i would use decimals and round off. approximations will change the value a bit.

but what is sqrt?
do you mean what the definition of a square root is?

...and the circumference of the two end is still 6 right?
no, the circumference is the same as the width of the cylinder

for the last one...will it work if i have the radius=5 a=6 and each side be 8? with a height of 8.8 and it be right for the volume?
no it would not. plug in these values into the volume equation, you won't get 196 cm cubed

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