Science:Math Exam Resources/Courses/MATH104/December 2016/Question 12/Solution 1

Let's begin with diagram:

Let $x\geq 0$ be the side length of the square which we cut out from the four corner. Indeed, $x$ becomes the height of the resulting box, The bottom of the box is also a square with the side length $12-2x\geq 0$ as we can see in the diagram. Then, the volume of the box is

$$V=x(12-2x)^{2}$$

Since $x>0$ and $12-2x>0$ is required, the domain of the function $V$ is $0\leq x\leq 6$. To find the largest volume (which is the global maximum of $V$ on $[0,6]$, we first find the derivative of $V$; using the product rule and the chain rule,

$${\begin{aligned}{\frac {dV}{dx}}&=(12-2x)^{2}+x\cdot 2(12-2x)(-2)\\&=(12-2x)[(12-2x)-4x]\\&=(12-2x)(12-6x)=12(6-x)(2-x)\end{aligned}}$$

Setting ${\textstyle {\frac {dV}{dx}}=0}$ gives us the critical numbers: ${\textstyle x=6}$ and ${\textstyle x=2}$. Since we have a closed interval, we can test the critical numbers and the 2 endpoints.

$${\begin{aligned}V(0)&=0\\V(2)&=2\cdot 8^{2}\\V(6)&=0\end{aligned}}$$

From which we see there is global maximum when ${\textstyle x=2}$. The height of the resulting box will be ${\textstyle x=2}$, while the length and width will be ${\textstyle 12-2x=8}$, therefore the dimensions are $\color {blue}8\times 8\times 2$.