Answer:
The dimensions of the rectangle are:
[tex]x \approx 46.87 \:m[/tex]
[tex]y \approx 46.87 \:m[/tex]
Step-by-step explanation:
Let x be the length and y be the width of the rectangle.
According with the graph the area is [tex]A=x\cdot y=2197[/tex] and its perimeter is [tex]P=2x+2y[/tex]
Solving the area function for y gives us
[tex]y=\frac{2197}{x}[/tex]
Plug this into the perimeter function
[tex]P(x)=2x+2(\frac{2197}{x})\\\\P(x)=2x+\frac{4394}{x}[/tex]
Because we want to minimize the perimeter function, find the derivative and set it equal to zero to locate the critical points.
[tex]\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}[/tex]
[tex]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[/tex]
[tex]\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}[/tex]
Because a width cannot be negative we only take as a valid solution:
[tex]x=13\sqrt{13}[/tex]
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 [tex]f''(x_0)>0[/tex], then f has a local minimum at [tex]x_0[/tex]
The second derivative is
[tex]\frac{d}{dx}P= 2-\frac{4394}{x^2}\\\\\frac{d^2}{dx^2}P=\frac{8788}{x^3}[/tex]
Plug [tex]x=13\sqrt{13}[/tex] into the second derivative
[tex]\frac{8788}{\left(13\sqrt{13}\right)^3}=\frac{4\sqrt{13}}{169}\approx0.085[/tex]
Because [tex]P''(13\sqrt{13})>0[/tex], [tex]13\sqrt{13}[/tex] is a local minimum.
The dimensions of the rectangle are:
[tex]x = 13\sqrt{13}\approx 46.87 \:m[/tex]
[tex]y=\frac{2197}{13\sqrt{13}}=13\sqrt{13}\approx 46.87 \:m[/tex]
The dimensions of a rectangle with an area 2,197 m² whose perimeter is as small as possible are 46.9m by 46.9m
The area of a rectangle is expressed as A = xy
Perimeter of the rectangle = 2(x + y)
Since A = 2,197m²
xy = 2197..............1
2(x+y) = Minimum ....................2
From 1, x = 2197/y
Substitute into equation 2:
2(2197/y + y) = minimum
4394/y + 2y = P(y)
If the perimeter is at minimum, hence dP/dy = 0
-4394/y² + 2 = 0
-4394/y² = -2
2y² = 4394
y² = 2197
y =√2197
y = 46.9m
Since xy = 2197
x = 2197/y
x = 2197/46.9
x = 46.9m
Hence the dimensions of a rectangle with an area 2,197 m² whose perimeter is as small as possible are 46.9m by 46.9m
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