Answer :

The dimensions would be 29 by 29.

To maximize area and minimize perimeter, we make the dimensions as close to equilateral as possible.  

Dividing the perimeter by the number of sides, we have

116/4 = 29

This means that both length and width can be 29.
abidemiokin

The dimensions of a rectangle with perimeter 116 m whose area is as large as possible is 29m by 29m

The formula for calculating the perimeter of a rectangle is expressed as:

P = 2(L+W)

If the perimeter is 116 m, then;

116 = 2(L + W)

58 = L + W .........1

The area of the rectangle is expressed as:

A = LW......... 2

From  equation 1;

L = 58 - W, substitute into 2:

A = (58 - W) W

A = 58W - W^2

If the  area is as large as possible, then dA/dW = 0

dA/dW = 58 - 2W = 0

2w = 58

w = 58/2

w = 29 m

Since 58 = L + W, hence l = 29m

The dimensions of a rectangle with perimeter 116 m whose area is as large as possible is 29m by 29m

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