(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant 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.

3. 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?

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# Homework Help: Optimization Problem/application of the derivative

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