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- Mar 1st 2009, 11:20 AMAlphaRockMax/Min Of Area/Volume
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- Mar 1st 2009, 01:29 PMarpitagarwal82
let length and width of bas be x (its square so both are equal)

and height be y.

so volume $\displaystyle x^2y = 2$

so$\displaystyle y = 10/x^2$

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

area of top =$\displaystyle x^2$

so cost

$\displaystyle C = 10(x^2 + 4xy) + 20x^2$

$\displaystyle = 30x^2 + 40xy$

$\displaystyle = 30x^2 + 400/x$ (since $\displaystyle y = 10/x^2)$

for minimum cost

dC/dt = 0

solve for x now.

then solve for y ($\displaystyle y= 10/x^2$)