An open-top box is to be made by cutting small congruent squares from the corners of a 12-in.by-12-in. sheet of tin and bending up the sides .How large should the squares cut from the corners be to make the box hold as much as possible?
solution:
I start with a picture .In the figure , the corner square are x in. on a side . The volume of the box is a function of this variable:
$V(x)=4x(12-2x)+(12-2x)^2=48x-8x^2+144-48x+4x^2=144-4x^2$
then
$\frac{dV}{dx}=-8x$
$\Longrightarrow \frac{dV}{dx}=0 \Longrightarrow x=0$
$\Longrightarrow V(0)=144$
The maximum volume is 144 in.$^3$.The cutout squares should be 0 in. on a side.
But this way is wrong,why?
Right way:
I start with a picture .In the figure,the corner square are x in. on a side.The volume of the box is a function of this variable:
$V(x)=x(12-2x)^2=144x-48x^2+4x^3$
then
$\frac{dV}{dx}=144-96x+12x^2=12(12-8x+x^2=12(2-x)(6-x)$
Since the sides of the sheet of tin are only 12 in. long,$x\leq 6$ and the domain of $V$ is the interval $0\leq x\leq 6$ .
Endpoint values:$V(0)=0, V(6)=0$
Critical-point value:$V(2)=128$ .
The maximum volume is 128 in.$^3$.The cutout squares should be 2 in. on a side.