1. ## Integration (?) question

a cuboid has the square base of side $\displaystyle xcm$ and height $\displaystyle hcm$. its volume is $\displaystyle 120cm^3$.

i) find $\displaystyle h$ in terms of $\displaystyle x$. Hence show that the surface area, $\displaystyle Acm^2$, of the cuboid is given by $\displaystyle A = 2x^2 + \frac{480}{x}$ .

ii) find $\displaystyle \frac{dA}{dx}$ and $\displaystyle \frac{d^2A}{dx^2}$

iii) Hence find the value of $\displaystyle x$ which gives the minimum of surface area. Find also the value of the surface area in this case.

Thanks alot guys, as always your amazing :P

2. Originally Posted by coyoteflare
a cuboid has the square base of side $\displaystyle xcm$ and height $\displaystyle hcm$. its volume is $\displaystyle 120cm^3$.

i) find $\displaystyle h$ in terms of $\displaystyle x$. Hence show that the surface area, $\displaystyle Acm^2$, of the cuboid is given by $\displaystyle A = 2x^2 + \frac{480}{x}$ .

ii) find $\displaystyle \frac{dA}{dx}$ and $\displaystyle \frac{d^2A}{dx^2}$

iii) Hence find the value of $\displaystyle x$ which gives the minimum of surface area. Find also the value of the surface area in this case.
$\displaystyle V = x^2h = 120$

solve for h ...

$\displaystyle h = \frac{120}{x^2}$

surface area is (top+bottom) + (4 sides) ...

$\displaystyle A = 2x^2 + 4xh$

sub in for h ...

$\displaystyle A = 2x^2 + 4x\left(\frac{120}{x^2}\right)$

$\displaystyle A = 2x^2 + \frac{480}{x}$

... try the rest of the problem yourself.