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Eight segments measuring $3,5,6,9,11,13,15,$ and $16$ inches are used to construct a rectangle. In square inches, what is the largest possible area of the rectangle?

I found the average side length of a side of the rectangle to be $\dfrac{78}{4} = 19.5$ since we are looking for the rectangle with maximal area it must be square like, but I don't see how this helps. Also there doesn't seem to be a systematic way of solving this.

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    $\begingroup$ It helps because you can get as close as possible to that with integers. $19=3+16=6+13$, $20=5+15=9+11$ $\endgroup$ Commented Dec 11, 2015 at 14:27

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Among the rectangles with the same perimeter, the one with the maximal area is the square.

So you have to ask yourself how close to a square you can get with those segments. First of all, the sides need to be pairwise equal. Note that the average segment is $9.75$. So let's try adding numbers whose sum is close to $2\cdot 9.75$, say it is between $18$ and $20$.
$$16+3 = 6+13 = 19$$ $$15+5 = 11+9 = 20$$ is the only working combination ($15+3 = 13+5 = 18$ but $6,9,11,16$ do not add up to two equal numbers)

So that is your answer.

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    $\begingroup$ Once you have found this $19\times20$ rectangle, you know you have the answer, because $n \times (39-n) < 19\times20$ for any $n < 19$. $\endgroup$
    – David K
    Commented Dec 11, 2015 at 14:55
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All indicated to me that the only possibility of construct a rectangle with ALL these eight lenghts is with the case $16+3=6+13$ and $15+5=11+9$. Miscalculation unless, there is no alternative.

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  • $\begingroup$ You can also get a (smaller) rectangle with $16+3+5=11+13$ and $15=6+9$. $\endgroup$ Commented Dec 12, 2015 at 1:06
  • $\begingroup$ Oh! I have played just with two numbers for side. Sorry. Thank you for your rectificatory comment. $\endgroup$
    – Piquito
    Commented Dec 13, 2015 at 12:17

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