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What is the formula to find the optimum width and the height of given units that would fit in an area? For example : I have 100 units that need to be placed in a rectangle that is 300 in height and 400 in width. How do I get the optimum width and height of a unit so that all 100 units can fit in the area?

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  • $\begingroup$ Well the longest distance line segment entirely in the rectangle is the diagonal. The diagonal cuts the rectangle into two right triangles. $\endgroup$ – fleablood Nov 21 '15 at 0:52
  • $\begingroup$ You may want different units in your example. 100 < 300 so any shape will do. $\endgroup$ – fleablood Nov 21 '15 at 0:55
  • $\begingroup$ I made some clarifications. $\endgroup$ – rastacide Nov 21 '15 at 0:56
  • $\begingroup$ A mere suggestion: don't ask "what is the formula".Math is more interesting than that. Try to understand. $\endgroup$ – leonbloy Nov 21 '15 at 1:08
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Your space has an area of $$300\times400=120,000$$ This is the same as the combined area of all 100 units. Therefore, you have $$120,000=100\times\text{Area of 1 unit}$$ Therefore, $$\text{Area of 1 unit}=\frac{120,000}{100}=1,200$$ Now, you have to figure out one thing: What am I optimizing the height and with in relation to? You know that $$\text{height}\times\text{width}=1,200$$ but you need to add another constraint to optimize the height and width. In this case, you want to optimize the perimeter (as you said in a now-deleted comment). You have $$P=2(\text{width}+\text{height})$$ You then can write $$\text{height}=\frac{1,200}{\text{width}}$$ and substitute that in to find $$P=2\left(\text{width}+\frac{1,200}{\text{width}}\right)$$ All you have to do is find the extrema of this function within the range of acceptable values (for example, the function admits 1,000 as a possible value for the width, but that clearly isn't a valid solution, because one unit would be wider than the total original area!).


Putting it in a simpler way, if $h$ is the height, $w$ is the width, $A$ is the area, and $P$ is the perimeter, then $$A=hw,\quad P=2w+2h$$ If you know either the perimeter or the area, you can optimize the other one.

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The most efficient rectangle with maximum area per perimeter is a perfect square. (Basic calculus.) So if 1/4 of the units is less than the smaller side of the rectangle, that is best. The square will slide right in.

If on the other hand you can't do a perfect square because the smaller side of the containing rectangle is too small, The second best thing is a rectangle with the sides as near in length as possible. So do a rectangle with one side equal to the smaller side of the rectangle. That's as close to a square as you can get.

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So assuming 25 units is less than 300 inches. Put this into a 25 by 25 unit square. That gives you area of 625 square units.

If 25 units is more the 300 inches this is impossible. So put it into a rectangle that 300 inches by (100 units - 600 inches)/2. That's the best you can do.

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