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This math problem popped up in my textbook:

A strip of metal is 24cm wide. This strip of metal is to be used to form a length of guttering for a house. The guttering will be open at the top so water can flow through. What are the dimensions of the cross section of the guttering if it is to hold the maximum volume of water?

I solved it, and got the correct answer of Width = 12cm, Depth = 6cm (therefore you will have to fold up 6cm of metal on each side).

However, what I cant understand is why a square wouldn't give maximum volume. It is often considered to be a given that to maximise area, a rectangle of equal length sides (i.e. A square) should be used. However, why is it that in this problem a rectangle is the optimum shape?

mvw
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Luke
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2 Answers2

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Intuitively, it is because you don't need material to cover the top of the gutter, so width costs half as much as height. This should cause the solution to be wider than a square. A clever approach is to note that if you make a copy of the gutter, invert it, and put it on top of your gutter, you have the classic problem of maximizing the area of a rectangle for a given perimeter. As you say, this is solved by a square, so the gutter problem should be twice as wide as it is high. All this is a bit of handwaving, and mvw's answer is spot on.

Ross Millikan
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Just considering rectangular cross sections, then $$ A(w) = w \frac{24 - w}{2} \quad (w \in [0, 24]) $$ At the borders of the interval we get $A=0$.

For local extrema we need $$ 0 = A'(w) = 12 - w $$ so at $w = 12$ is a critical point.

As $A''(w) = -1 < 0$ it is a local maximum.

Indeed $A(12) = 72 > A(8) = 64$, so the square solution is not optimal.

If we had considered closed profiles we had the objective function: $$ A(w) = w \frac{24 - 2 w}{2} \quad (w \in [0, 12]) $$ Note: That the domain changed as well.

Then again $$ 0 = A'(w) = 12 - 2 w = 2(6 - w) $$ so at $w = 6$ is a critical point. And this now is the square profile.

Conclusion: The open upper side makes the difference.

gutter problems (Large version)

The open gutter objective function (green) and the closed gutter objective function (blue).
The red dots show the square cross section solution each.

mvw
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  • Looks like this is answering the textbook problem. But I thought the OP said ey already solved the textbook problem? – Don Hatch Oct 25 '16 at 18:57
  • I tried to explain that the square is optimal if all four sides are involved, but here one side is not and a non-square rectangle is optimal. – mvw Oct 26 '16 at 17:44