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## Homework Statement

A wire 6 meters long is cut into twelve pieces, eight of one length and four of another. These pieces are welded together at right angles to form the frame of a box with a square base.

a) Where should the cuts be made to maximize the volume of the box?

b) Where should the cuts be made to maximize the are of the box?

## Homework Equations

The derivative in relation to volume and area.

Perimeter=6meters

Perimeter=8x+4y=6meters

x=(4y-6)/8

y=(6-8x)/4

A(x)=4xy + 2x^2

V(x)=x^2y

A'(x)=8x + 4y

V'(x)=2xy + x^2

where some 'x' and some 'y' are of some meters and less than 6 meters.

## The Attempt at a Solution

So, I understand that setting the derivative of a function equal to zero of a function can allow someone to find the maximum and minimum of that function which is very practical in many areas, however, I can't seem to find the solution to this problem. It appears that the solution doesn't exist, but I must be doing something wrong.

Here's what I did for a)

a)

since V'(x)=0 when volume is optimized, then

0=2xy + x^2

y=(6-8x)/4

0=(2x(6-8x)/4) + x^2

-x^2=(6x/2) - (4x^2)

3x^2=(6x/2)

x= 1 meter

oh I just solved it,

but if x=1meter, then y=((6-8(1))/4)meters=-1/2meters, but how is this quantity a negative? This is where I get stuck. I must be doing something wrong.

b)

since A'(x)=0 where area is optimized

0=8x + 4y

y=(6-8x)/4

8x=-4(6-8x)/4

32x=-24+32x

x=0? o.0??? can someone show me what I'm doing wrong?