Find the dimensions of a rectangle with area 1000 $m}^{2$ whose perimeter is as small as possible.

Margie Marx
2021-12-16
Answered

Find the dimensions of a rectangle with area 1000 $m}^{2$ whose perimeter is as small as possible.

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lenkiklisg7

Answered 2021-12-17
Author has **29** answers

Let it be: A-area, x width im miles and y-length in miles.

$A=x\cdot y$

We have$A=1000$ , so

$1000=x\cdot y$

$y=\frac{1000}{x}$

Let it be Perimeter - P

$P=2x+2y$

$P=2(x+y)$

Substitute$y=\frac{1000}{x}$ in equation $P=2(x+y)$

$P=2x+\frac{2000}{x}$

Compute${P}^{\prime}\left(x\right)$

$P}^{\prime}\left(x\right)=2-\frac{2000}{{x}^{2}$

Find critical points.

$2-\frac{2000}{{x}^{2}}=0$

${c}^{2}=1000$

$c=10\sqrt{10}$

Compute$P{}^{\u2033}\left(x\right)$ :

$P{}^{\u2033}\left(x\right)=\frac{4000}{{x}^{3}}$

$P{}^{\u2033}\left(10\sqrt{10}\right)=\frac{4000}{100000\sqrt{10}}>0$

By applying The Second Derivative Test, we have that$x=-\frac{1}{4}$ is a minimum.

Find y:

We have:

$y=\frac{1000}{10\sqrt{10}}$

Rationalize:

$y=\frac{1000}{10\sqrt{10}}\cdot \frac{10\sqrt{10}}{10\sqrt{10}}$

$y=10\sqrt{10}$

$P=40\sqrt{10}$

Result:$x=10\sqrt{10}\text{}mi\le s$

$y=10\sqrt{10}\text{}mi\le s$

We have

Let it be Perimeter - P

Substitute

Compute

Find critical points.

Compute

By applying The Second Derivative Test, we have that

Find y:

We have:

Rationalize:

Result:

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