 # Find the dimensions of a rectangle with area 1000 m^2 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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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
$P=2x+2y$
$P=2\left(x+y\right)$
Substitute $y=\frac{1000}{x}$ in equation $P=2\left(x+y\right)$
$P=2x+\frac{2000}{x}$
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 $P{}^{″}\left(x\right)$:
$P{}^{″}\left(x\right)=\frac{4000}{{x}^{3}}$
$P{}^{″}\left(10\sqrt{10}\right)=\frac{4000}{100000\sqrt{10}}>0$
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: