Question

Calc 3, max min values without Lagrange multiplier.

Answer 1

@ganeshie8

Answer 2

A rectangular box with no top and two parallel partitions (see the figure below) must hold a volume of 64cubic inches. Find the dimensions that will require the least amount of material.

This is the question, I did all of it but when I got to checking the dimensions I got = 0, meaning the test fails.

Answer 3

here's a drawing, I guess I can put the steps, but since the test fails, would that means my dimensions are wrong then haha?

Answer 4

\[V = 64 \text{inches}^3~~~, V = xyz,~~~\text{Surface Area} S = 4xz+2yz+xy\]

Answer 5

\[V = 64 = xyz \implies z = \frac{ 64 }{ xy }\]
\[S(x,y) = 4x(64/xy)+2y(64/xy)+xy,~~~x>0,~~y>0 \]
\[S(x,y) = \frac{ 256 }{ y }+\frac{ 128 }{ x }+xy\]

\[\frac{ \partial s }{ \partial x } = \frac{ -128 }{ x^2 }+y= 0,~~~S_{xx} = \frac{ 128 }{ x^3 },~~~S_{xy} = 1\]

\[\frac{ \partial S }{ \partial y } = \frac{ -256 }{ y^2 }+x=0,~~~S_{yy} = \frac{ 256 }{ y^3 }\]

Answer 6

Letting it = 0, etc, I got the dimensions (4,8,2)

but when I check for the dimensions are at a minimum
\[D = S_{xx}(4,8) \times S_{yy}(4,8)-[S_{xy}(4,8)]^2 \implies \frac{ 128 }{ 4^3 } \times \frac{ 256 }{ 8^3 }-1 = 0 \]

Answer 7

Guess I could check with Lagrange eh haha.

Answer 8

so it is a degenerate case ?

Answer 9

since you have only one critical point, cant u conclude it is a minimum ?

Answer 10

That's what I thought at first, but I wasn't entirely sure because it was like you mentioned degenerate haha.

Answer 11

you don't have many critical points to decide which point is what

Answer 12

when u get a degenerate case, it just means u need to decide the type of critical point using some othe rmeans

Answer 13

Ah alright, cool, the critical point is a minimum then.

Answer 14

Since it's >0

Answer 15

Thanks Ganeshie :D

Answer 16

it is a minimum because your intuition+wolframm says so
http://www.wolframalpha.com/input/?i=min+4xz%2B2yz%2Bxy%2C+xyz%3D64

for your professor, you can simply say
"cannot conclude using second derivative test"

Answer 17

unless you're given other tools to analyze the degenerate case...

Answer 18

Nope, nothing to deal with it. Probably in the next course, thanks again!

Answer 19

Now, I'm curious, are there other ways to deal with it?

Answer 20

idk, looks higher order derivatives helps in deciding the nature of critical point
read this
http://math.stackexchange.com/questions/721432/inconclusive-second-derivative-test

Answer 21

this is interesting
http://en.wikipedia.org/wiki/Higher-order_derivative_test

Answer 22

Very nice stuff, I'm going to give this a read, thanks.

Answer 23

that wiki link works for single variable functions
similar stuff must be there for multivariable functions also... need to check

Answer 24

Hey @ganeshie8 I'm curious about this, and I'm going over this again, and when I check for the dimensions I see it's not a degenerate case, does it matter which dimensions you pick, x,y or x,z, etc?

Answer 25

If I do yz (height, width) I get a local min. Mhm.

Answer 26

Sorry should be zx

Answer 27

If I use xz instead of xy for my dimensions to verify it's a local min, this way I get it not being a degenerate case, so would that be allowable?