# Calculus 3 Help needed-Minimizing

• Jul 21st 2009, 11:46 AM
Reskal
Calculus 3 Help needed-Minimizing
I'm self teaching myself multivariable calculus and have gotten stuck on problems involving minimizing and maximizing. The two problems that I am currently stuck on are as follows.

1)You are creating a retangular box with a volume of 32000cm^3. Find dimensions that minimize the material used in the construction of this box.

2) Find three numbers whose sum add to 100 and whose product is a minimum.

When I work both problems my answers come out as maximums and not minimums, assistance will be appreciated very much.

Thank you
• Jul 21st 2009, 12:33 PM
arbolis
Hi,
Quote:

When I work both problems my answers come out as maximums and not minimums
Did you use the Lagrange multipliers method?
• Jul 21st 2009, 12:40 PM
Jester
Quote:

Originally Posted by Reskal
I'm self teaching myself multivariable calculus and have gotten stuck on problems involving minimizing and maximizing. The two problems that I am currently stuck on are as follows.

1)You are creating a retangular box with a volume of 32000cm^3. Find dimensions that minimize the material used in the construction of this box.

2) Find three numbers whose sum add to 100 and whose product is a minimum.

When I work both problems my answers come out as maximums and not minimums, assistance will be appreciated very much.

Thank you

I got the first to work out - show us what you tried. For the second, you'll need to try the boundary values (is zero allowed or do the numbers have to be positive?)
• Jul 23rd 2009, 11:43 AM
Reskal
The numbers must be positive for both problems and cannot be zero.

For the first problem I set up the two equations

xyz=32000
2xz+2yz+xy=f(x,y,z)

Solved for z=3200/xy
and then set that in the second equation to get

64000/y+64000/x+xy=f(x,y,z)

taking the gradient I get (-64000/x^2+y, -64000/y^2+x)

When I began to do the second derivative test and solved for fxx I realized that my answer would yield a maximum and was stuck there.

(langrange multipliers is not until the next section so I doubt I am supposed to use those, will these start making these problems easier?)

For the second problem

I set the two equations to be x+y+z=100 and xyz=f(x,y,z)

solved for z=100-x-y and subed that into the second equation to get

xy(100-x-y)

took the gradient of that to get

fx=100y-2xy-y^2
fy=100x-x^2-2xy

and got stuck after that.

Thanks all!
• Jul 23rd 2009, 12:09 PM
Jester
Quote:

Originally Posted by Reskal
The numbers must be positive for both problems and cannot be zero.

For the first problem I set up the two equations

xyz=32000
2xz+2yz+xy=f(x,y,z)

Solved for z=3200/xy
and then set that in the second equation to get

64000/y+64000/x+xy=f(x,y,z)

taking the gradient I get (-64000/x^2+y, -64000/y^2+x)

When I began to do the second derivative test and solved for fxx I realized that my answer would yield a maximum and was stuck there.

(langrange multipliers is not until the next section so I doubt I am supposed to use those, will these start making these problems easier?)

For the second problem

I set the two equations to be x+y+z=100 and xyz=f(x,y,z)

solved for z=100-x-y and subed that into the second equation to get

xy(100-x-y)

took the gradient of that to get

fx=100y-2xy-y^2
fy=100x-x^2-2xy

and got stuck after that.

Thanks all!

First is the box and open top (you didn't mention that). Second, factor $f_x$ and $f_y$ and look at case where each is zero.