optimization problems:container for wooden matches

[wind]

Hi, I am having trouble with this question, can some one pleas check over my work? Thanks

A container for wooden matches consists of an open-topped box (to contain the matches) that slides into an outer box, open at both ends. The length if the boxes are fixed at 10cm to match the length of the matches. The outer box is designed so that one side completely overlaps for gluing.

a) If the outer box is made form a sheet of cardboard that is 16cm by 10cm, what dimensions for the outer box will maximize the capacity?

b) What size should the sheet of cardboard be, to make the inner box in this case?


a) 2L + 2h = 16
h = 16-2L/2
h = 8-L

v=Lwh
v=(2L)(10)(8-L)
v=160L-20L^2

v'=160-40L
0=160-40L
-160=-40L
4=L

therefore L=4 is a possible max or min
v''= -40

when its like this do I use the chart? and do i say the test fails, or is that when the second derivitive equals zero?

h=8-4
h=4

but that would mean that the face is a suqare..and it does not say that in the question...

b) the inner box is the same as the biggerone,but with no top right? so

2L+2h-L=16
L=16/2h
L=8h

the volume is the same so

320=(8h)(10)(2h)
320=160h^2
2=h^2
root2=h

...I think I did that wrong, can some one help me? Thanks

 

[skeeter]

since one end "overlaps" for gluing ...

16 = 3x + 2y where x = box height and y = box width

V = 10xy

get the volume equation in terms of one variable and maximize.

 

[wind]

ok thanks, skeeter

so

16=3h+2w
(16-2w)/3=h

v=10hw
v=10[(16-2w)/3]w
v=160-20w/3]w
v=160w-20w^2/3

v'=160-40w/9
0=160-40w
4=w

therefore 4 is a possible max or min

v''=-40/81

what do I do? do I use the chart? Did I do something wrong?

so if i found that then I would have the width, and i would plug that into (16-2w)/3=h to find the height. right?

then how should I do the second part? the same thing except use 2h + 2w = 16?

thanks

 

[skeeter]

ok thanks, skeeter

so

16=3h+2w
(16-2w)/3=h

v=10hw
v=10[(16-2w)/3]w
v=160-20w/3]w
v=160w-20w^2/3 v = (160w - 20w²)/3

v'=160-40w/9 v' = (160 - 40w)/3
0=160-40w
4=w

therefore 4 is a possible max or min

v''=-40/81 v" = -40/3 ... this tells you what about the volume function and the nature of the extrema at w = 4?

what do I do? do I use the chart? Did I do something wrong? no

so if i found that then I would have the width, and i would plug that into (16-2w)/3=h to find the height. right? yes

then how should I do the second part? the same thing except use 2h + 2w = 16?no ... 2h + 2w = inner box sheet length ... it doesn't have to overlap.
thanks

 

[wind]

no ... 2h + 2w = inner box sheet length ... it doesn't have to overlap.

is the volume the same?

v=10hw
32/30=10hw
16/15h=w


2h + 2w = 10
2h + 2(16/15h)=10
62/15h=10
h=2.42

..I don't get it

 

[skeeter]

I forgot that the inside tray has no top. perhaps this will correctly explain it.

you've maximized the volume of the box ...

L = 10
w = 4
h = 8/3

the inside tray is a box w/ no top. it can be made with a single sheet where a square region of dimensions h by h in each corner is cut out and the resulting flaps are folded up.

length of the sheet would have to be 10 + 2(8/3) = 46/3
width of the sheet would have to be 4 + 2(8/3) = 28/3

 

[wind]

oh, I get it now , thanks skeeter!