Question

Some airlines have restrictions on the size of items of luggage that passengers are allowed to take with them. Suppose that one has a rule that the sum of the length, width and height of any piece of luggage must be less than or equal to 204 cm. A passenger wants to take a box of the maximum allowable volume. If the length and width are to be equal, what should the dimensions be?

length = width =___________
height =________
In this case, what is the volume?
volume =___________
(for each, include units)

f the length is be twice the width, what should the dimensions be?
length =

Answer 1

make a cube

Answer 2

Continuation

width =
height =
In this case, what is the volume?
volume =
(for each, include units)

Answer 3

ok

Answer 4

@aum

Answer 5

@ganeshie8

Answer 6

after the box. whats next?

Answer 7

@satellite73

Answer 8

Volume V = L * W * H
L + W + H = 204
L = W (given)

V = W * W * H = W^2H
2W + H = 204

V = W^2(204 - 2W)
Maximize V

Answer 9

@aum i found that
length = width =68cm
height= 68

could help me get the volume?

Answer 10

Yes. V = 68^3 cm^3

Answer 11

can you help me with the second part too?

Answer 12

\(\large V = 314432 ~~cm^3\)

Answer 13

Follow the same procedure as the first.

Answer 14

the second part is If the length is be twice the width, what should the dimensions be?
length =
width =
height =
In this case, what is the volume?
volume =
(for each, include units)

Answer 15

Same procedure as the first:

Volume V = L * W * H
L + W + H = 204

L = 2*W (given)
V = 2W * W * H = 2 * W^2 * H
3W + H = 204
V = 2W^2(204 - 3W)
Maximize V

Answer 16

@aum i found that height is 68cm

Answer 17

Yes. What about L and W?

Answer 18

im kinda lost there

Answer 19

How did you find H without finding either L or W?

Answer 20

what i did was

i got this
2x+y = 204 then i solved for y and i got
y= 204-2x (then i multiplied by x^2)
and i got y = 204x^2 -2x^3 (then took the derivative)
and i got 408x-6x^2 then solved for x
i got x = 0 and 68, i discarded x = 0 and thats how i found height
@aum

Answer 21

that leaves me to think that there is 134cm to spare

Answer 22

and that is all i got

Answer 23

@aum you get it now?

Answer 24

The above looks like answer for the first problem.

Answer 25

If instead of L and W you want to use x and y that is fine.
Then looks t my steps for the second part of the problem, replace L by y and W by x.
There is still H for height and you can use z if you want.

Answer 26

Let
length = y
width = x
height = z

Volume = xyz --- (1)

x + y + z = 204 ---- (2)
length is twice the width means y = 2x. Put this in (1) and (2)

(1) becomes: Volume = xyz = x(2x)z = 2x^2z --- (3)
(2) become: x + 2x + z = 204; 3x + z = 204; z = 204 - 3x ---- (4)

Put (4) in (3): Volume V = 2x^2(204-3x) = 408x^2 - 6x^3
Maximize volume V
dV/dx = 408*2*x - 18x^2 = 0. x(816 - 18x) = 0; x = 816/18 = 45.33 cm.

Put x = 45.33 in (4) and find z. Then put x and z in (2) and find y.

Answer 27

im lost afte the volume equation

Answer 28

got it thanks again @aum