Question

a rectangular box has a square base. The volume of the box is 256 cubic centimeters. The material for the base and the top of the box cost 0.10 per square centimenter and the material for the sides cost 0.05 per square centimenter. Find the dimensions of the box that minimize the cost of producing box.

Answer 1

no wonder you really hate math

Answer 2

LOOL calculus kills me

Answer 3

we can do this if you want

Answer 4

omg you know how to do that :)

Answer 5

yeah it will just take a few minutes. ok base is a square label the side x

Answer 6

ok

Answer 7

so area of the bottom and top are both \[x^2\]

Answer 8

they cost .10 per square whatever and you will have two of them (top and bottom) so total cost for the top and bottom is \[.10\times 2x^2\]

Answer 9

or if you prefer \[.2x^2\] these decimals are killing me. ok we continue

Answer 10

now the volume is 256 which we get by multiplying area of base times height.

Answer 11

the base has area \[x^2\] so we know \[x^2h=256\] i.e. \[h=\frac{256}{h}\]

Answer 12

sorry
\[h=\frac{256}{x}\]

Answer 13

still with me?

Answer 14

Yes :)

Answer 15

ok we have 4 sides. the base of each side is x and the height is
\[h=\frac{256}{x^2}\] so their area is each
\[x\times \frac{256}{x^2}=\frac{256}{x}\]

Answer 16

there are four of them so multiply by 4 to get the total area of the sides all together

Answer 17

i get 1024/x

Answer 18

so we multiply that by .05

Answer 19

and these each cost .05 (dollars, whatever)

Answer 20

yes

Answer 21

i get 51.2/x

Answer 22

me too

Answer 23

now i forget what the other stuff cost.

Answer 24

oh .2x^2

Answer 25

ok so total cost is\[ .2x^2+\frac{51.2}{x}\]

Answer 26

and this is the damned thing you have to find the minimum of. now for the easy part. take the derivative, find the critical points etc. ready?

Answer 27

yeah lol

Answer 28

actually it is not that bad.

Answer 29

\[\frac{d}{dx} .2x^2+\frac{51.2}{x}=.4x-\frac{51.2}{x^2}\]

Answer 30

now set = 0 and solve
\[.4x-\frac{51.2}{x^2}=0\]
\[\frac{.4x^3-51.2}{x^3}=0\]
\[.4x^3-51.2=0\]

Answer 31

\[x^3=\frac{51.2}{.4}=128\]

Answer 32

and finally.....
\[x=\sqrt[3]{128}\]

Answer 33

5.04 rounded. i am exhausted!

Answer 34

omg
I think I love u haha

Answer 35

please please dont tell me the answer in the back of the book is 12!

Answer 36

thanx radar!

Answer 37

looooll no

Answer 38

is that all or do you have a pile of these wonders to work out?

Answer 39

I have one more lool but I dont wanna bother you again

Answer 40

go ahead if i get tired i will leave

Answer 41

sqroot (x+y)=1+((x^2)(y^2))

Answer 42

ahh implicit diff?

Answer 43

your teacher must hate you

Answer 44

find the y'
and compute the value at (0,1)

Answer 45

ok we differentiate wrt x

Answer 46

remembering that when we take the derivative of y we get y'

Answer 47

here goes
\[\sqrt{x+y}=1+x^2y^2\]

Answer 48

\[\frac{1}{2\sqrt{x+y}}\times y'=2xy^2+x^22yy'\]

Answer 49

left hand side was because the derivative of root x is one over two root x, and the y' by the chain rule. left hand side was the product rule. first the derivative of x^2 is 2x and next the derivative of y^2 is 2y y' again by the chain rule

Answer 50

now we solve for y'
\[\frac{y'}{\sqrt{x+y}}-2x^2yy'=2xy^2\]
\[y'(\frac{1}{\sqrt{x+y}}-2x^2y)=2xy^2\]
\[y'=\frac{2xy^2}{\frac{1}{\sqrt{x+y}}-2x^2y}\]

Answer 51

i forgot the 2 outside the square root sign. sorry. it is there.

Answer 52

i will simplify that if you like

Answer 53

where are the sqroot lol

Answer 54

in the bottom

Answer 55

here it is simplified

Answer 56

\[\frac{4xy^2\sqrt{x+y}-1}{1-4x^2y\sqrt{x+y}}\]

Answer 57

now plugging in (0,1) should be nice and easy so i will let you do that. have fun!

Answer 58

loool thx a LOT good nite:)