Calculus I - Optimization

Section 4.8 : Optimization

8. We have a piece of cardboard that is 50 cm by 20 cm and we are going to cut out the corners and fold up the sides to form a box. Determine the height of the box that will give a maximum volume.

Two sketches side by side with an arrow pointing from the left sketch to the right one. On the left is the flat piece of cardboard, 50 cm wide and 20 cm tall, with a small square cut out of each of the four corners and dashed fold lines marked. On the right is the open-topped box that results when the four sides are folded up.

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The first step is to do a quick sketch of the problem.

$Two sketches side by side with an arrow pointing from the left sketch to the right one. On the left is the flat piece of cardboard with a square of side $h$ cut out of each of the four corners, leaving a width of $50 - 2h$ across the bottom and a height of $20 - 2h$ up the side. On the right is the open topped box that results when the sides are folded up, with dimensions $50 - 2h$ by $20 - 2h$ by $h$.$

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As with the problem like this in the notes the constraint is really the size of the box and that has been taken into account in the figure so all we need to do is set up the volume equation that we want to maximize.

\[V\left( h \right) = h\left( {50 - 2h} \right)\left( {20 - 2h} \right) = 4{h^3} - 140{h^2} + 1000h\]

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Finding the critical point(s) for this shouldn’t be too difficult at this point so here is that work,

\[V'\left( h \right) = 12{h^2} - 280h + 1000\hspace{0.75in}h = \frac{35 \pm 5\sqrt {19}}{3} = 4.4018,\,\,18.9315\]

From the figure above, we can see that the limits on $h$ must be $h = 0$ and $h = 10$ (the largest $h$ could be is ½ the smaller side). Note that neither of these really make physical sense but they do provide limits on $h$.

So, we must have $0 \le h \le 10$ and this eliminates the second critical point and so the only critical point we need to worry about is $h = 4.4018$

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Because we have limits on $h$ we can quickly check to see if we have maximum by plugging in the volume function.

\[V\left( 0 \right) = 0\hspace{0.75in}V\left( {4.4018} \right) = 2030.34\hspace{0.75in}V\left( {10} \right) = 0\]

So, we can see then that the height of the box will have to be $\require{bbox} \bbox[2pt,border:1px solid black]{{h = 4.4018}}$ in order to get a maximum volume.