Question

Let's construct a box whose base length is 3 times its base width and will have no lid. We only have 64 m2 of material to use. Determine the dimensions of the box that will give the largest volume.


___width?
___height?
___length?

What is the volume of the box?

ANY HELP ON HOW TO START THIS PLEASE??

Answer 1

we know that the surface area has to equal 64m^2 right?

Answer 2

our dimensions can be l,w,h

were l = 3w

Answer 3

surface area = 3w(w) + 2(hw) + 2(3wh) = 64

Answer 4

volume of the box = 3w(w)h

Answer 5

64 = 3w^2 +2wh +6wh

64 = 3w^2 +8wh

solve for "h"

64 - 3w^2 = 8wh

64-3w^2
-------- = h
8w

Answer 6

use this "value" of h in the formula for the volume:

V = 3w^2 [(64-3w^2)/8w]

Answer 7

V = (192w^2 - 9w^4)/8w ; now simplify

Answer 8

V = (192x -9w^3)/8

find the derivative of this Volume function to determine critical numbers

Answer 9

8(192-27w^2)
-------------
8(8)

192-27w^2
---------- solve for zero
8

192 - 27w^2 = 0

gonna be tricky for me without pencil and paper :)

Answer 10

got it..

3(64-9w^2) = 0

64-9w^2 = 0

(8+3w)(8-3w) = 0

w = -3/8 or w = 3/8

since a negative width is meaningless; lets use w = 3/8

Answer 11

recall that l = 3w:

l = 3(3/8) = 9/8

Answer 12

64-3w^2
-------- = h
8w

64 - 3(3/8)^2
------------ = h
8(3/8)

64 - (27/64)
------------ = h
3

Answer 13

4069/64
-------- = h
3

4069/192 = h

might need to reduce that....