1. ## Optimization Problem

Hi
i am having trouble getting the correct answer on the following question:
A box open at the top has a volume of $10m^3$. Find the dimensions which give minimum surface area.

(1) $V=xyw=10$

(2) $A=2xy+2yw+xw$

$w=\frac{10}{xy}$

sub w into (2) equation

$A=2xy+2y(\frac{10}{xy})+x({\frac{10}{xy})$
$=2xy+\frac{20}{x}+\frac{10}{y}$

$\frac{\partial A}{\partial x} = 2y-\frac{20}{x^2}$

$\frac{\partial A}{\partial x} = 0$
(3) $2y-\frac{20}{x^2} = 0$

$\frac{\partial A}{\partial y} = 2x-\frac{10}{y^2}$

$\frac{\partial A}{\partial y} = 0$
(4) $2x-\frac{10}{y^2} = 0$

make x the subject of equation (4)
$2x=\frac{10}{y^2}$
$x=5y^2$

sub back into equation (3)
$2y-\frac{20}{5y^2} = 0$
$2y-4y^2 = 0$
$2y(1-2y) = 0$

therefore

$y=0$or $\frac{1}{2}$

0 is unrealistic so don't consider it.

$x = \frac{5}{4}$

$w= 16$

my answer $\frac{1}{2}$m , $\frac{5}{4}$m, 16m

answer is cubic root 20m, cubic root 20m, 0.5 cubic root 20m

P.S

2. 2x = 10/y^2, So x = 5/y^2

2y - 20/5y^2 = 0

2y - 4/y^2 = 0

y^3 = 2.

Check the simplifications.