# Thread: Calculus 3 Help needed-Minimizing

1. ## 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

2. Hi,
When I work both problems my answers come out as maximums and not minimums
Did you use the Lagrange multipliers method?

3. 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?)

4. 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!

5. 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.