Let's begin with diagram:
Let
be the side length of the square which we cut out from the four corner. Indeed,
becomes the height of the resulting box,
The bottom of the box is also a square with the side length
as we can see in the diagram. Then, the volume of the box is
![{\displaystyle V=x(12-2x)^{2}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2a3c66d5477b5ec0b4fd2c47aa49d705a34cfc61)
Since
and
is required, the domain of the function
is
. To find the largest volume (which is the global maximum of
on
, we first find the derivative of
; using the product rule and the chain rule,
![{\displaystyle {\begin{aligned}{\frac {dV}{dx}}&=(12-2x)^{2}+x\cdot 2(12-2x)(-2)\\&=(12-2x)[(12-2x)-4x]\\&=(12-2x)(12-6x)=12(6-x)(2-x)\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f537b62ec3ce83eadc2e196b13b6d725abb2937f)
Setting
gives us the critical numbers:
and
. Since we have a closed interval, we can test the critical numbers and the 2 endpoints.
![{\displaystyle {\begin{aligned}V(0)&=0\\V(2)&=2\cdot 8^{2}\\V(6)&=0\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4477c33dc905e673b144291b6aa14f86cd7df6e7)
From which we see there is global maximum when
. The height of the resulting box will be
, while the length and width will be
, therefore the dimensions are
.