Respuesta :
The smallest perimeter and dimensions of rectangle having area of 169 [tex]inches^{2}[/tex] is 52 inches , length 16 inches and breadth also 16 inches.
Given that the area of rectangle is 169 [tex]inches^{2}[/tex].
We have to find the smallest perimeter and dimensions of the rectangle.
let the length of rectangle be x inches and breadth be y inches.
We know that area of rectangle is the product of its length and breadth. So,
Area=xy
169=xy
y=169/x
Perimeter of rectangle =2(x+y)
Put the value of y=169/x in the perimeter.
P=2(x+169/x)
P=2x+338/x
Differentiate both sides with respect to x.
dP/dx=2-338/[tex]x^{2}[/tex]
Again differentiate it with respect to x.
[tex]d^{2} P/dP^{2}[/tex]=676/[tex]x^{3}[/tex]
Because the second derivative is greater than 0 so the perimeter is minimum.
Put dP/dx=0 to get the value of x.
dP/dx=0
2-338/[tex]x^{2}[/tex]=0
2=338/[tex]x^{2}[/tex]
[tex]x^{2}[/tex]=169
x=[tex]\sqrt{169}[/tex]
x=13
Put the value of x in y=169/x
y=169/13
=13
To calculate the perimeter, put the values of x and y in P=2(x+y)
P=2(13+13)
=2*26
=52 inches.
Hence The smallest perimeter and dimensions of rectangle having area of 169 [tex]inches^{2}[/tex] is 52 inches , length 16 inches and breadth also 16 inches.
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