Cfrgtkky

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 &= (xcdot y) \
frac{d}{dx}1000&=frac{d}{dx}(x cdot y) \
0 &=frac{d}{dx}((x)prime(y)+(y)prime(x)) \
0 &=y+xfrac{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!

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 $xy=1000implies x=dfrac{1000}{y}$
$$P=2left(dfrac{1000}{y}right)+2y$$
$$P^{prime}=-dfrac{2000}{y^2}+2$$

The above equation does not exist if $y=0$

$$-dfrac{2000}{y^2}+2=0$$
$$y^2=1000$$
$$y=pmsqrt{1000}$$

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 $x=dfrac{1000}{sqrt{1000}}impliessqrt{1000}$ and width $y=sqrt{1000}$

So, we can say that the perimeter of the rectangle is minimized which is actually a square.

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