maximizing volume

jessica87689

New member
Joined
Dec 2, 2008
Messages
3
A 12 by 32 in rectangular piece of cardboard is used to construct and open box by cutting identical squares from each corner and then folding up the sides the size of the squares that must be removed to maximize the volume of the open box.

okay, ive gotten this far:
V= (12 -2x) (32 -2X) (X)
v= (384 -88x + 4X^2)x
V ' = (384x -088x^2 + 4x^3)
V ' = 4( 3x^2 - 44x+ 96 )

i know you need to find an optimization problem and then the restraints, but i dont know how to take this one any further.
 
thanks for the reply, i did that and came up with:

4(3x^2-44x+96)=0
4 (3x-8)(x-12)
(12x-32)(X-12)
x = 8/3 and x = 12.

my only worry is that i went wrong somewhere in the problem because in comparison to the size of the box, 8/3 in X 12 in doesnt seem quite right.
 
jessica87689 said:
thanks for the reply, i did that and came up with:

4(3x^2-44x+96)=0
4 (3x-8)(x-12)
(12x-32)(X-12)
x = 8/3 and x = 12.

my only worry is that i went wrong somewhere in the problem because in comparison to the size of the box, 8/3 in X 12 in doesnt seem quite right.

You are forgetting - what does 'x' mean in your problem.

That is why it is a good idea to write out the definition of your variable at the beginning of your problem(or could it be somebody gave you those equations - and you did not know).

x = Side of the square cut out from the corner

Now interpret your result....
 
Top