# Max/Min Of Area/Volume

• March 1st 2009, 12:20 PM
AlphaRock
Max/Min Of Area/Volume
(deleted)
• March 1st 2009, 02:29 PM
arpitagarwal82
let length and width of bas be x (its square so both are equal)

and height be y.

so volume $x^2y = 2$
so $y = 10/x^2$

area of sides and bottom = $x^2 + 4xy$
area of top = $x^2$

so cost
$C = 10(x^2 + 4xy) + 20x^2$
$= 30x^2 + 40xy$
$= 30x^2 + 400/x$ (since $y = 10/x^2)$

for minimum cost
dC/dt = 0

solve for x now.
then solve for y ( $y= 10/x^2$)