So I have a textbook question I have no idea how to do –
An open-top box is to be made from a square piece of metal with $20$cm long sides by cutting equal area squares from the corners of the sheet of metal and then folding up the sides. Determine the side of the square which is to be cut out so that the volume of the box will be maximized.
I don't have any idea where to start. Any suggestions?
Best Answer
Let $x$ be the length of the side of the corner squares that need to be cut.
Then each side of the resulting box will be $20 - 2x$ cm long, which means the area of the base will be $(20 - 2x)^2$, and the height of the box will be $x$.
Let $V(x)$ represent the volume of the box as a function of $x$. Volume is area of base $\times$ height:
$$V(x) = x(20 - 2x)^2 = 4x(10 - x)^2 = 4x(100 - 20x + x^2)= 400x- 80x^2 + 4x^3$$
Now, we find $V'(x)$, set equal to zero to find critical points:
$$\begin{align} V'(x) & = 400 - 160 x + 12x^2 = 0\\ \\ & = 4(100 - 40x + 3x^2) = 0 \\ \\ & = 4(3x - 10)(x - 10) = 0 \\ \\ & \iff (3x - 10) = 0 \;\text{or}\;(x - 10) = 0 \\ \\ & \iff x = \dfrac {10}{3} \;\text{or} \;x = 10\end{align}$$
All that's left, now, is to determine which of the "zeros" is a maximum: one of the critical points will give the length $x$ of a side of the square to be cut that results in a box of maximum volume. (And the other yields a box with NO volume: i.e., minimizes volume.)
NOTE:
I trust you can determine that $\bf x = \dfrac {10}{3}$ cm is a maximum (i.e., maximized volume, and there is no need for taking the second derivative to do this!), and which is the minimum. Indeed, if $x = 10$, we'd cut four $10 \times 10$ corners, and there'd be nothing left to form a box! So $x = 10$ yields no box, or a box with the absolute minimum of volume: $0$cm$^3$.