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Find the smallest perimeter and the dimensions for a rectangle with an area of 169 in\(^2\).

**Part 1:** The smallest perimeter for a rectangle with an area of 169 in\(^2\) is _____ in.

Answer :

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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