Find the maximum and minimum volumes of a rectangular box whose surface area is 1300 cm2 and whose total edge length is 200 cm.

Answer 1

This is an exercise in algebra, I think.

max volume will be when it is a square box .
min volume is when one of the lengths is zero.

min volume: h=0, 4w+4L=200 and w=L

max volume: Vmax=h^3
when 200=4w+4L+4h=12h or h=50/3 cm

test these by picking one dimension, and varying it.

Answer 2

no you have to use lagrange multipliers to do it and i figured out all five equations but i can not solve any of them

Answer 3

f(x,y,z)=xyz & 2xy+2xz+2yz=1300 & 4x+4y+4z=200

FIRST, simplify constraints:
g(x,y,z)=xy+yz+xz=650
h(x,y,z)=x+y+z=50

THEN, take gradient vectors:
for ▽f=<yz, xz, xy>=λ▽g+μ▽h
=<λ(y+z), λ(x+z), λ(x+y)>+<μ, μ, μ>

SO:
yz=λ(y+z)+μ (#1)
xz=λ(x+z)+μ (#2)
xy=λ(x+y)+μ (#3)

(#1)-(#2) implies λ=z while (#1)-(#3) implies λ=y
Therefore λ=y=z however x can't also be equal b/c a cube violates the constraints

Then using that info to solve we get:
xy+yz+xz=650 -->x(2λ)+λ^2=650
--> x=(650-λ^2)/(2λ)
x+y+z=50 -------> x=50-2λ

Setting these equal: 50-2λ=(650-λ^2)/(2λ)
And Solving we get: λ=(5/3)*(10±sqrt(22)

And since x=50-2λ and y=z=λ, we can just plug this back into f(x,y,z) to get f(x,y,z)'s extreme values.

The Max: (1/27)*(42500+5500sqrt(22))
The Min: (1/27)*(42500-5500sqrt(22))

Answer 4

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