Question

A square based, box-shaped shipping crate is designed to have a volume of 16 ft^3. The material used to make the base costs twice as much (per ft^2) as the material in the sides, and the material used to make the top costs half as much (per ft^2) as the material in the sides. What are the dimension of the crate that minimize the cost of materials?

Answer 1

@amistre64 please help I'm struggling

Answer 2

@robtobey

Answer 3

A square based, box-shaped shipping crate is designed to have a volume of 16 ft^3.

b*b*h = 16

The material used to make the base costs twice as much (per ft^2) as the material in the sides,

4sides of area bh

1base of area bb

and the material used to make the top costs half as much (per ft^2) as the material in the sides.

1top of area bb

Total surface area?: 4bh + 2bb


hmmm

Answer 4

its a rate of change problem i can see that

Answer 5

so far i have got to exactly where you are at

Answer 6

We need a total Cost equation to minimise

Answer 7

lets say S = cost for sides
B = cost of base = 2S
T = cost of top = S/2

Answer 8

Cost = 4bhS + bb2S + bbS/2

Answer 9

that should be doable, but a gotta go ..... for a min :)

Answer 10

ok I'll keep trying to figure this out

Answer 11

lets implicit this, and recall that the S is a constant price

Answer 12

ok

Answer 13

Cost = 4S bh + 2S b^2 + S b^2/2

Cost' = 4S(b'h+bh') + 2S(2b b') + S(2b b')/2

Cost' = 4S(b'h+bh') + 4S bb' + S bb'

Recall that we are given:

V = bbh

V' = 2bb' h+b^2 h' ; and since the volume remains constant, V' = 0

0 = 2bh b'+b^2 h'

-2bh b' = b^2 h'

-2b' h = bh'; might be able to use this to simplify stuff


To minimize costs, we want Cost' to be 0

0 = 4S(b'h+bh') + 4S bb' + S bb'

0 = 4Sb'h + 4Sbh' + 5S bb'

-4S b'h = 4Sbh' + 5S bb' ; -4b' h = 2bh'

2S bh' = 4Sbh' + 5S bb' ; divide out the S

2bh' = 4bh' + 5 bb'

-2bh' = 5 bb' ; divide of the b

-2h' = 5b'

looks to be getting someplace .... what was the question again?

Answer 14

oh yeah, find bh that minimize costs

Answer 15

h' = -5b'/2

h = -5b/2 .... gotta think if thats useful or not now :)

Answer 16

if we let b' = 1 .....

0 = 4S(b'h+bh') + 4S bb' + S bb'

0 = 4S(h+bh') + 4S b + S b

and h' = -5/2, while h=-5b/2


0 = 4(-5b/2-5b/2) + 4b + b

0 = 4(-5b) + 4b + b

0 = b(-20+4+1) grrr, missing something im sure

Answer 17

0 = 4(-5b/2-5b^2/2) + 4b + b
^^
5b^2 , not 5b ;)

Answer 18

0 = -20b/2-20b^2/2 + 5b

0 = -10b - 10b^2 + 5b

0 = -10b^2 -5b

0 = 5b(2b + 1)

b = 0, or b = -1/2; lol .... i wonder if 1/2 is suitable

Answer 19

says that the dimensions should be \[\frac{ 4 }{ \sqrt[3]{5} },\frac{ 4 }{ \sqrt[3]{5} },5^{\frac{ 2 }{ 3 }}\]

Answer 20

thats the answer key?

Answer 21

yessir

Answer 22

on paper i was getting something like b = 4/sqrt(2) :)

Answer 23

we can assume the cost to be an arbitrary value like 1 for simplicity

Answer 24

well that's extremely close. Much closer than anything i would've gotten without your help

Answer 25

Cost = 1(4bh) + 2(b^2) + b^2/2
sides base top

Cost = 4bh + 5b^2/2 is a simpler cost function to me

16 = b^2h, such that h=16/b^2

Answer 26

that's what i started out doing

Answer 27

Cost = 4b16/b^2 + 5b^2/2

Cost = 64b^(-1) + 5/2 b^2

Cost' = -64b^(-2) + 5b ; min max is when Cost' = 0

5b = 64b^-2

5b^3 = 64

b = cbrt(64/5)

Answer 28

not real sure why i didnt simplify it beforehand :)

Answer 29

\[b=\sqrt[3]{\frac{64}{5}}=\frac{4}{\sqrt[3]{5}}\]

Answer 30

how did you get the -64b^-2

Answer 31

do you agree that the cost function is: Cost = 4bh + 5b^2/2

Answer 32

do you agree that the constraint for the volume is: 16 = b^2h, giving us h = 16b^(-2) ?

Answer 33

oh yea

Answer 34

if so then:

Cost = 4bh + 5b^2/2

Cost = 4b(16b^(-2)) + 5b^2/2

Cost = 64b^(-1) + 5b^2/2

Answer 35

the rest is just a power derivative

Answer 36

ok see i didn't realize we were taking the derivative yet

Answer 37

had to take it at sometime ;) but yeah, it worked up simpler than i started .... getting old bites ;)

Answer 38

lol. Hey is there a way to save conversations or something so i could come back to this later?

Answer 39

A solution using Mathematica for the calculations is attached.

Answer 40

thanks a lot man means alot that you took the time out to help as well