How do you find the dimensions of a rectangle whose area is 100 square meters and whose perimeter is a minimum?

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Answer 1 · 2014-10-08T02:06:06

Let $x$ $x$ and $y$ $y$ be the base and the height of the rectangle, respectively.

Since the area is 100 $m^{2}$ $m^2$ ,

$x y = 100 \Rightarrow y = \frac{100}{x}$ $xy=100 Rightarrow y=100/x$

The perimeter $P$ $P$ can be expressed as

$P = 2 (x + y) = 2 (x + \frac{100}{x})$ $P=2(x+y)=2(x+100/x)$

So, we want to minimize $P (x)$ $P(x)$ on $(0, \infty)$ $(0,infty)$ .

By taking the derivative,

$P'(x)=2(1-100/x^2)=0 Rightarrowx=pm10$

$x=10$ is the only critical value on $(0,infty)$

$y=100/10=10$

By testing some sample values,

$P'(1)<0 Rightarrow P(x)$ is decreasing on $(0,10]$ .

$P'(11)>0 Rightarrow P(x)$ is increasing on $[10,infty)$

Therefore, $P(10)$ is the minimum

I hope that this was helpful.

Hence, the dimensions are $10\times10$ .