Find the dimensions of a rectangle with perimeter 108 m whose area is as large as possible. (If both values are the same number, enter it into both blanks.) m (smaller value) m (larger value)

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tqiu

Answer:

27 by 27

Step-by-step explanation:

Let the sides be x and y. The problem is essentially asking:

Given 2(x+y)=108, maximize xy.

We know that x+y=54. By the Arithmetic Mean - Geometric Mean inequality, we can see that [tex]\frac{x+y}{2} \ge \sqrt{xy[/tex]. Substituting in x+y=54, we get [tex]27\ge\sqrt{xy}[/tex], meaning that [tex]729 \ge xy[/tex]. Equality will only be obtained when x=y (in this case it will generate the maximum for xy), so setting x = y, we can see that x = y = 27. Hence, 27 is the answer you are looking for.

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