Optimization Problem. Find Smallest Perimeter of a Rectangle Given Area.
Emily Wong 
QUESTION
Find the dimensions of a rectangle with area $1000$m$^2$ whose perimeter is as small as possible.
MY WORK
I think we are solving for $\frac{dy}{dx}$:\begin{align*} P &= (2x+2y) \\ A &= (x\cdot y) \\ \frac{d}{dx}1000&=\frac{d}{dx}(x \cdot y) \\ 0 &=\frac{d}{dx}((x)\prime(y)+(y)\prime(x)) \\ 0 &=y+x\frac{dy}{dx}\\ \frac{dy}{dx} &=\frac{-y}{x} \end{align*}
I didn't use the perimeter formula. I am not sure where I messed up. I think probably somewhere in my setup for the equation. If someone could take a look at my work and point me in the right direction it would be greatly appreciated!
$\endgroup$ 02 Answers
$\begingroup$Take the sides of the rectangle as length $x$ and width $y$.
Given area is $xy=1000$
We need to minimize the perimeter of the rectangle. So, $P=2x+2y$
To minimize this, we need to differentiate $P$. Since, we have two different variables in the perimeter, let us bring it to one variable.
We have $xy=1000\implies x=\dfrac{1000}{y}$$$P=2\left(\dfrac{1000}{y}\right)+2y$$$$P^{\prime}=-\dfrac{2000}{y^2}+2$$
The above equation does not exist if $y=0$
$$-\dfrac{2000}{y^2}+2=0$$$$y^2=1000$$$$y=\pm\sqrt{1000}$$
Since the length cannot be negative we can ignore $-\sqrt{1000}$
Therefore, $y=\sqrt{1000}$ is the critical number. By plotting we can see that $y=\sqrt{1000}$ do not correspond to be local minimum, which means the perimeter of the width is minimized.
We have length $x=\dfrac{1000}{\sqrt{1000}}\implies\sqrt{1000}$ and width $y=\sqrt{1000}$
So, we can say that the perimeter of the rectangle is minimized which is actually a square.
$\endgroup$ 2 $\begingroup$For minimum perimeter, put $\frac{dP}{dx} = 0$ $$\Longrightarrow 1+\frac{dy}{dx} = 0$$$$\Longrightarrow 1+(-\frac{y}{x}) = 0$$$$\Longrightarrow \frac{x-y}{x}=0$$$$\Longrightarrow x=y$$So perimeter will be minimum when the rectangle is a square.
Hope it is helpful
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