Optimization Problem. Find Smallest Perimeter of a Rectangle Given Area.

Question

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!

Answer 1

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.

Answer 2

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