A box with no top. Minimize cardboard used. How???

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Kristy

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#10 Directions:
A box with a square base and open top must have a volume of 32,000 cm^3.
Fnd the dimensions of the box that minimize the amount of material used.

Ok it would have a bottom that would be Length x Width,
2 sides that would be Length x Height
2 sides that would be Width x Height

Volume = L*W*H =32,000 cm 3

Surface area (without top) = L*W + 2 L * H + 2 W x H = f (something)
Then I know I’m supposed to take the derivative, but how do I do that with three variables.
 
Since the base is square you only have two variables.


The volume is \(\displaystyle \L\\V=x^{2}y=32,000\).........[1]

The surface area is \(\displaystyle \L\\S=x^{2}+4xy\).........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.
 
Kristy said:
#10 Directions:
A box with a square base and open top must have a volume of 32,000 cm^3.
Fnd the dimensions of the box that minimize the amount of material used.
Click to expand...

Thanks. I kinda missed the "square base" part too. That would have made it more obvious. Oops. As soon as I saw you x^2 I wondered why you could do that, and then....okay its square. :(

Okay, makes much more sense now.
 
galactus said:
Since the base is square you only have two variables.


The volume is \(\displaystyle \L\\V=x^{2}y=32,000\).........[1]

The surface area is \(\displaystyle \L\\S=x^{2}+4xy\).........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.
Click to expand...



solve [1] for y
The volume is \(\displaystyle \L\\V=x^{2}y=32,000\).........[1]

\(\displaystyle \L\\y = \frac{32,000}{x^{2}}\)



and sub into [2]
The surface area is \(\displaystyle \L\\S=x^{2}+4xy\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+4x(\frac{32,000}{x^{2}})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+4(\frac{32,000}{x})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+(\frac{128,000}{x})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+128,000(\frac{1}{x})\).........[2]













in progress...wow this tex stuff is hard! Almost as haard as the math!
 
Yep. Now sub that into the surface area equation. It'll be entirely in terms of x and you can differentiate and all that.
 
Sorry I kept editing my post because of trying to make sure I was doing the tex right.


Kristy said:
galactus said:
Since the base is square you only have two variables.


The volume is \(\displaystyle \L\\V=x^{2}y=32,000\).........[1]

The surface area is \(\displaystyle \L\\S=x^{2}+4xy\).........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.
Click to expand...



solve [1] for y
The volume is \(\displaystyle \L\\V=x^{2}y=32,000\).........[1]

\(\displaystyle \L\\y = \frac{32,000}{x^{2}}\)



and sub into [2]
The surface area is \(\displaystyle \L\\S=x^{2}+4xy\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+4x(\frac{32,000}{x^{2}})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+4(\frac{32,000}{x})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+(\frac{128,000}{x})\).........[2]

The surface area is \(\displaystyle \L\\S=x^{2}+128,000(\frac{1}{x})\).........[2]


in progress...wow this tex stuff is hard! Almost as haard as the math!
Click to expand...


The surface area is \(\displaystyle \L\\S=x^{2}+128,000(x^{-1})\).........[2]


Then do the differentiating, set to 0 thing and solve for x

S'(x) = 2x + 128,000 * (-1)\(\displaystyle x^{-2}\)

S'(x) = 2x + 128,000 *\(\displaystyle \frac{-1}{x^{2}}\)

0 = 2x + 128,000 *\(\displaystyle \frac{-1}{x^{2}}\)

\(\displaystyle \frac{128,000}{x^{2}} = \frac{2x}{1}\)

\(\displaystyle 2x^{3}=128,000\)

divide both sides by 2

\(\displaystyle x^{3}=64,000\)

take the cube root of both sides
(is there a negative cube root too or is that just with squre roots?

x=4,000

So that means the bottom is 4,000X4,000

the height is y and is

\(\displaystyle \L\\y = \frac{32,000}{x^{2}}\)


=\(\displaystyle \L\\y = \frac{32,000}{4^{2}}\)

y = 2000

So the dimensions are 4000 length and with and 2000 height.

Does that seem right?
 
You're doing good until S'(x)=2x.

What about differentiating \(\displaystyle \L\\\frac{128000}{x}\)?.

After you have it, set to 0 and solve for x.
 
Thanks for the encouragement. I think I got it now, not sure though. I guess a box could be twice as long and wide and not very tall, I'm not really sure how to check this one though.
 
You get:

\(\displaystyle \L\\S'(x)=2x-\frac{128000}{x^{2}}\)

Set to 0 and solve for x:

\(\displaystyle \L\\2x-\frac{128000}{x^{2}}=0\)

\(\displaystyle \L\\2x^{3}=128000\)

\(\displaystyle x=40, \;\ and \;\ y=20\)

See?. That wasn't hard. :D It that what you got?.
 
galactus said:
You get:

\(\displaystyle \L\\S'(x)=2x-\frac{128000}{x^{2}}\)

Set to 0 and solve for x:

\(\displaystyle \L\\2x-\frac{128000}{x^{2}}=0\)

\(\displaystyle \L\\2x^{3}=128000\)

\(\displaystyle x=40, \;\ and \;\ y=20\)

See?. That wasn't hard. :D It that what you got?.
Click to expand...

I got extra 0's for some reason I don't know.
I got x to be 4000 and y 2000 see my post above.
I do'nt know what I did to multiply everythhing by 100
 
The cube root of 64000 isn't 4000, it's 40. That's why you're off by a multiple of 100.
 
galactus said:
The cube root of 64000 isn't 4000, it's 40. That's why you're off by a multiple of 100.
Click to expand...

Oops. What I did was take the cube root of 64, found it to be 4, and then added the 000 so it was 4000.

So the dimensions are x = 40 length and width
height y = 20

?
 
Kristy said:
(is there a negative cube root too or is that just with squre roots?
Click to expand...

Positive numbers can have only positive cube-roots.

Negative numbers can have only negative cube-roots.
 
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