Optimisation question

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?

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$^3$.

• very nice write-up +1 Sep 8 '13 at 0:10

Let $x$ be the size of the square which is to be cut. We have then that the size of the bottom of the box will be once folded : $$(20-2x)^2 = \text{Area of the square which is the bottom of the box}$$ Then we multiply this with the height of your box to get the volume, the height will be equal to $x$ since you just fold the side up. We have then : $$x*(20-2x)^2 = \text{Area} *\text{Height} =\text{Volume of the folded box}$$

And now you want to maximize it. So its first derivative must be equal to $0$ and its second derivative must be smaller than 0 in this point to be a local maximum. Let's control this.

1st derivative : $$\frac{d}{dx} \phantom{t}(x (20-2 x)^2) = 4 (3 x^2-40 x+100)$$ And 2nd derivative : $$\frac{d^2}{dx^2} \phantom{t} (x (20-2 x)^2) = 8 (3 x-20)$$ Let's solve it : first derivative must be equal to $0$ to be an extrema. $$4 (3 x^2-40 x+100)=0 \Rightarrow x=10/3 \text{ or } x=10$$ And we have then :

• if $x=10$ : $8 (3 x-20) = 80$
• if $x=10/3$ : $8 (3 x-20) = -80$ The second derivative must be smaller than $0$ to be a maximal point. (If it's bigger than 0, it's a minimum.)

And thus we have found that $$x=10/3$$ is a local maximum for the volume of your box.

Voilà !

Note: as said by amWhy in his own note, if you are doing it as a physician or as an engineer, you can bypass the second derivative test since it's not possible in case of $x=10$, but as a Mathematician, you can't go like that IMO and should always test your extrema points thoroughly to know whether they are maximal, minimal or inflection's points. ;-)