Question

For an open-topped rectangular box, with a square base x by x cm and height h cm, find the dimensions giving the minimum surface area, given that the volume is 13 cm^3.

***write EXACT answers***
x=_____cm
h=_____cm

please explain! Thank you!! :)

Answer 1

Well this is definitely...odd, lets see what we can do here...

Answer 2

Well for a rectangular prism...the volume is going to be LxWxH

L and W are the same 'x' ...so lets write

\[\large V = x^2 \times h\]

Answer 3

yeah haha definitely odd :P

okay i'm following so far:)

Answer 4

so then you go \[13=x^2 * h\] ?

Answer 5

Well we're going to want the area of the bottom...plus the 4 sides of the box

(4 sides because we have the base...and then the 4 sides (open top))

Answer 6

okay... so multiply by 4?

Answer 7

So first lets express that Volume equation in terms of h

if V = x^2 x H

H = V/x^2

Answer 8

okay:)

Answer 9

Alright...so the area of the bottom...is just the area of a square..so Length x Width....which we know is x^2 right?

Answer 10

mhmm:)

Answer 11

But then we need to ADD the area of the 4 sides...

Well each of the 4 sides have an area of (x times h)

So...altogether we have

A = x^2 + (4xh)

We know h = v/x^2 so

\[\large A = x^2 + 4x(\frac{V}{x^2})\]

Answer 12

Alright yeah I think I'm coming out with the right answer lol..okay have faith we're almost done haha

Answer 13

hahaha of course :P

okay, so i followed that part:)

Answer 14

Alright.. well we can see that an 'x' can be canceled there on the right

\[\large A = x^2 + \frac{4V}{x}\]

follow that?

Answer 15

yes :)

Answer 16

Awesome...

okay here's the only annoying part "minimize the surface area" -_-

haha not too bad! We just need to take the derivative of this function and set it equal to 0

Answer 17

okay yeah this stuff is annoying haha

ermm so

A' = 2x - 4V
----
x^2
?

not sure about that second term :/

Answer 18

How are you not sure!? That's completely correct! :) lol

Answer 19

oh!! okay awesome!! yay :)

hehe so \[2x - \frac{ 4V }{ x^2 } = 0\] ?

Answer 20

are we solving for x?

Answer 21

Alright...so remember that is equal to 0

\[\large 2x - \frac{4V}{x^2} = 0\]

So we can do

\[\large 2x = \frac{4V}{x^2}\]

and yes solving for 'x' there

Answer 22

okay, so 2x^3 = 4V

x^3 = 2V ?

so x = cube root 2V ? ;/

Answer 23

Perfect again...haha you and your "unsure face" :P

Answer 24

haha woo!

and yeah.... it's appropriate though since i wasn't sure ahhahaa :P

Answer 25

So looky we have our 2 equations to solve

\[\huge x = \sqrt[3]{2V}\]
and
\[\huge h = \frac{V}{x^2}\]

so plug in your 13 for V and solve for 'x' and then you can solve for 'h'

Answer 26

okay:)

ermm so x= cube root 2(13)
=cube root 26
?? would that be considered the exact answer for x? :/

Answer 27

Indeed! they want to be PICKY and be EXACT! haha

Answer 28

ahh okay! :)

so x= cube root 26

and h=13/(cube root 26)^2

?? and would we just leave that like that?

Answer 29

Yeah...I mean that is correct

\[\huge x = \sqrt[3]{26}\]
\[\huge h = \frac{13}{\sqrt[3]{26}^2}\]

I 'm not sure if they want 'h' like that or like

\[\huge h = \frac{13}{26^{\frac{2}{3}}}\]

but yeah...either way ...you arrive at the correct solution

Answer 30

ahh okay:) yay!! :)

so that's all there is to this problem? :O

Answer 31

That's it..

and we can check obviously by doing

\[\huge V = x^2 \times h\]
\[\huge 13 = \sqrt[3]{26}^2 \times \frac{13}{\sqrt[3]{26}^2}\]
\[\huge 13 = \frac{13\times \cancel{26^{2/3}}}{\cancel{26^{2/3}}} = 13\]

hey looky there :D

Answer 32

haha yeah looky there! :)

awesome!! Thanks so much!! :D

this makes sense now!! :D

Answer 33

No problem :)