Respuesta :
Answer:
The dimensions of the rectangle are: [tex]6\sqrt{2},3\sqrt{2}[/tex]
The smallest perimeter therefore will be:[tex]18\sqrt{2}[/tex]
Step-by-step explanation:
Let the dimension of the rectangle be x and y
Area of the Rectangle = xy
Area=36 Inch Squared
Therefore: xy=36
[tex]x=\frac{36}{y}[/tex]
Perimeter of the Rectangle, P(x,y)=2(x+y)
Substituting [tex]x=\frac{36}{y}[/tex] into P(x,y)
[tex]P(y)=2(\frac{36}{y}+y)=\frac{72+y^2}{y}[/tex]
The minimum value of the perimeter occurs at the point where the derivative =0.
[tex]P'(y)=\frac{y^2-72}{y^2}[/tex]
[tex]Setting\:P'(y)=0\\\frac{y^2-72}{y^2}=0\\y^2-72=0\\y^2=72\\y=\sqrt{72}=6 \sqrt{2}[/tex]
Recall:
[tex]x=\frac{36}{y}\\=\frac{36}{6\sqrt{2} }\\x=3\sqrt{2}[/tex]
The dimensions of the rectangle are: [tex]6\sqrt{2},3\sqrt{2}[/tex]
The smallest perimeter therefore will be:
[tex]=2(6\sqrt{2}+3\sqrt{2}) \\=2(9\sqrt{2})\\=18\sqrt{2}[/tex]
