Question

rectangular container with no top is to have a volume of 8m^2. Find the dimensions that will minimize the amount of material used. This is in terms on xyz.

Answer 1

So this is what i did... xyz=8
z=8/xy
A= xy+2xz+2yz
then sub in the z

Answer 2

A(x,y) xy+16/y +16/x

I'm having probs with finding the values for x,y, z

Answer 3

Ax= y-16/x^2
Ay=x-16/y^2

Answer 4

How about considering f(x,y,z) instead?
f(x,y,z) = xy + 2xz + 2yz
and find
fx, fy, and fz ?

Answer 5

i assume you mean 8 m^3 right?

Answer 6

in my text it says m^2 but i think its a mistake

Answer 7

do you know lagrangian multipliers? .. what math class is this?

Answer 8

nope its 2nd part of calculus 1

Answer 9

integrals

Answer 10

so as @terenzreigns said, why not consider a general rectangle with dimensions x,y,z what is the volume of that?

Answer 11

and the surface area of that (surface area = amount of material used in this case)

Answer 12

isn't that just A= xy+2xz+2yz ?

Answer 13

yup.

Answer 14

so how can we minimize that while ensuring that x*y*z = 8?

Answer 15

find the zeros?

Answer 16

make z= 8/xy then sub them into A= xy+2xz+2yz

Answer 17

so that's one way you can do it. you can then find a minimum of that function.

In[1]:= (x*y + 2*x*z + 2*y*z) /. z -> 8/(x*y)

In[2]:= f = 16/x + 16/y + x y

Out[2]= 16/x + 16/y + x y

In[30]:= NSolve[D[f, x] == 0 && D[f, y] == 0, {x, y}, Reals]

Out[30]= {{x -> 2.51984, y -> 2.51984}}

then you solve for z = 8/(x*y) = 1.25992

this gives you a minimized surface area of 19.0488