Find the dimensions that will minimize the surface area of a Rectangle

[chwala]

Homework Statement

See attached.

Relevant Equations

##\nabla f=0##

My interest is on number 11.

In my approach;

##v= xyz##

##1000=xyz##

##z= \dfrac{1000}{xy}##

Surface area: ##f(x,y)= 2( xy+yz+xz)##

##f(x,y)= 2\left( xy+\dfrac{1000}{x} + \dfrac{1000}{y}\right)##

##f_{x} = 2y -\dfrac{2000}{x^2} = 0##

##f_{y} = 2x -\dfrac{2000}{y^2} = 0##

On solving the simultaneous, i have

##2xy^2 - 2x^2y=0, 2xy(y-x)=0##

##(x_1, y_1) = (0,0)## is a critical point but ##x,y ≠ 0## leaving us with

##y-x=0, ⇒ y=x## thus,

##2x^3 - 2000=0##

##x_{2}=10, ⇒ y_{2} =10## and therefore ##z=\dfrac{1000}{100} =10##

thus the dimensions are ##(x,y,z) = (10,10,10)##.

also,

##D (10,10)= \left[\dfrac{4000}{x^3} ⋅ \dfrac{4000}{y^3} - 2^2 \right]= 16-4=12>0## and ##f_{xx} (10,10) = 4>0## implying that ##f## has a local minimum at ##(10,10).##

For avoidance of doubt, ##D = f_{xx} ⋅f_{yy} - (f_{yy})^2##

Your wise counsel is welcome or any insight. Cheers guys.

 

[anuttarasammyak]

[tex]xyz=1000[/tex]
[tex]A=2(xy+yz+zx)=2000(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})[/tex]
In symmetry we expect
[tex]\frac{1}{x}=\frac{1}{y}=\frac{1}{z}=\frac{1}{1000^{1/3}}=\frac{1}{10}[/tex]
is the case we seek. A=600.　

[EDIT]
We can prove that
[tex] \sqrt[3]{abc} \leq \frac{a+b+c}{3} [/tex]