Question

Find the dimensions of a rectangle with area 2,197 m2 whose perimeter is as small as possible. (if both values are the same number, enter it into both blanks.) m (smaller value) m (larger value)

Answer 1

Answer:

The dimensions of the rectangle are:

$x \approx 46.87 \:m$

$y \approx 46.87 \:m$

Step-by-step explanation:

Let x be the length and y be the width of the rectangle.

According with the graph the area is $A=x\cdot y=2197$ and its perimeter is $P=2x+2y$

Solving the area function for y gives us

$y=\frac{2197}{x}$

Plug this into the perimeter function

$P(x)=2x+2(\frac{2197}{x})\\\\P(x)=2x+\frac{4394}{x}$

Because we want to minimize the perimeter function, find the derivative and set it equal to zero to locate the critical points.

$\frac{d}{dx}P=\frac{d}{dx}\left(2x+\frac{4394}{x}\right) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(\frac{4394}{x}\right)\\\\\frac{d}{dx}P=2-\frac{4394}{x^2}$

$2-\frac{4394}{x^2}=0\\2x^2-\frac{4394}{x^2}x^2=0\cdot \:x^2\\2x^2-4394=0\\2x^2-4394+4394=0+4394\\2x^2=4394\\x^2=2197$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{2197},\:x=-\sqrt{2197}\\x=13\sqrt{13},\:x=-13\sqrt{13}$

Because a width cannot be negative we only take as a valid solution:

$x=13\sqrt{13}$

After establishing the critical points of this function, the second derivative test uses the value of the second derivative at those points to determine whether such points are a local maximum or a local minimum.

If $f''(x_0)>0$, then f has a local minimum at $x_0$

The second derivative is

$\frac{d}{dx}P= 2-\frac{4394}{x^2}\\\\\frac{d^2}{dx^2}P=\frac{8788}{x^3}$

Plug $x=13\sqrt{13}$ into the second derivative

$\frac{8788}{\left(13\sqrt{13}\right)^3}=\frac{4\sqrt{13}}{169}\approx0.085$

Because $P''(13\sqrt{13})>0$, $13\sqrt{13}$ is a local minimum.

The dimensions of the rectangle are:

$x = 13\sqrt{13}\approx 46.87 \:m$

$y=\frac{2197}{13\sqrt{13}}=13\sqrt{13}\approx 46.87 \:m$

Answer 2

Let the dimensions of the rectangle be x and y .

And area is given by length times width.

So we will get

$A= xy$

$2197 = xy=> x = \frac{2197}{y}$

Let perimeter is represented byp. And perimeter is the sum of all sides, that is

$p = 2x+2y$

Substituting the value of x , we will get

$p =2*\frac{2197}{y} +2y = \frac{4394}{y}+2y$

Differentiating ,

$p' = -\frac{4394}{y^2}+2$

Differentiating again,

$p''=\frac{13182}{y^3} >0$

Since p''>0, so we will get maximum perimeter.

Now we need to set P' to 0 and solve for y, that is

$-\frac{4394}{y^2} +2=0$

$y^2 = 2197$

$y=46.9m$

So ,

$x = \frac{2197}{46.9}=46.9m$