# How Square tin with side "6a" units after bending have length of a side of the square is "a" units?

Suppose by using square tin sheet with side "6a units", make a topless box of maximum volume by cutting equal squares at the corners and removing them and then bending the tin so as to form the sides of the box. What I don't understand is that how the length of a side of the square becomes "a units."?

Let the side of the cut square be $x$. Then, the sides and height of the resultant cuboid will be $6a - 2x, 6a - 2x, x$.

Volume= $(6a - 2x)(6a - 2x) x = 4x (3a - x)^2$.

For maximum value of volume, $\frac{dv}{dx} = 0$.

$\Rightarrow 4(3a-x)^2 + 8x(3a-x)(-1) =0$

$\Rightarrow (3a-x)(3a-3x)=0$

Because $x=3a$ not possible as then two sides will have zero length, $x=a$ units.

• Thank you @Rohan for helping.. Dec 10 '16 at 8:31

By cutting equal squares of side $x$ at the corners and removing them we can construct a box of dimensions $6a-2x$, $6a-2x$, $x$ with $0\leq x\leq 3a$.

Hence by AGM-inequality its volume is bounded by $$V(x):=x(6a-2x)^2=2(2x)(3a-x)^2\leq 2\left(\frac{2x+(3a-x)+(3a-x)}{3}\right)^3=16a^3=V(a).$$ Moreover the equality holds iff $2x=3a-x$, that is $x=a$.

• @Zonnie You are welcome! Any further doubt? Dec 10 '16 at 8:39
• No.. That's so kind of you.. Dec 11 '16 at 7:03