Snacks will be provided in a box with a lid (made by removing squares from each corner of a rectangular piece of card and then folding up the sides)
You have a piece of cardboard that is 40cm by 40 cm – what dimensions would give the maximum volume?
This is how I attempted it
Let the length of the square to be cut off be x cm.
V be the Volume in cm3
Volume = L x B x H
$$L= 40 – 2x$$ $$B= 40 – 2x$$ $$H = x$$ So Volume $= x(40-2x)(40-2x)$ $$V = 4X^3 – 160x^2 + 1600x$$ Or $V= x^3 – 40x^2 + 400x$ $$V' = 3x^2 – 80x + 400$$ $$V'' = 6x – 80$$ Solve for turning points by putting $V ' = 0$
$$3x^ 2 – 80x + 400 = 0 $$
$$(3x - 20)(x – 20) = 0$$
$x= 20/3$ and $x = 20$
For Max Volume, $x = 20$ is a reasonable solution that can be apply and discard the other root i.e. $x = 20$
2nd derivative test $x = 20/3$:
$$V'' = 6(20/3) – 80= -20 < 0 $$ it will give Maximum Volume
$$V = (20/3)3 – 40(20/3)2 + 400(20/3)$$
$$V = 1186 \text{cm}^ 3$$
Have I done it right ?
$\endgroup$ 31 Answer
$\begingroup$$$\begin{align} l&=40-2x\\ b&=\frac{40-2x}{2}\\ &=20-x\\ h&=x\\ V&=lbh\\ &=(40-2x)(20-x)x\\ &=800x-80x^2+2x^3\\ V'&=800-160x+6x^2\\ &=2(x-20)(3x-20) \end{align}$$
By observation, the root at $x=20$ gives $b=0$ so $V=0$, so this clearly can't be the maximum.
That leaves $x=\frac{20}{3}$, where $V=3,555.5$.
For completeness, you should also check the boundaries - these may be the maximum within the domain even though they are not maximas. $x=20$ we have dealt with, $x=0$ falls over because $h=0$.
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