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]

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

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