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Tanisha has 54 m of fencing to build a three-sided fence around a rectangular plot of land that sits on a riverbank. (The fourth side of the enclosure would be the river.) The area of the land is 280 square meters. List each set of possible dimensions (length and width) of the field.

Possible dimensions #1: _ meters by _ meters.
Possible dimensions #2: _ meters by _ meters.

Respuesta :

Possible dimensions of the field are; 13.5 m by 20.74 m or 20.74 m by 13.5 m

How to maximize area?

Since it is a rectangular plot, let the opposite sides be x meter

Thus, the third side = (54 – 2x) m

Area (A) is given by;

A = (54 – 2x)x

A = 54x - 2x²

We will differentiate to get;

dA/dX = 54 - 4x

For a maximum Area we will differentiate and equate to zero i.e. dA/dx = 0. Thus;

54 – 4x = 0

x = 54/4

x = 13.5 m

Area is 280 m². Thus, the other possible dimension is;

w = 280/13.5

w = 20.74 m

Read more about Maximizing Area at; https://brainly.com/question/19819849

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Answer

Step-by-step Possible dimensions of the field are; 13.5 m by 20.74 m or 20.74 m by 13.5 mHow to maximize area?Since it is a rectangular plot, let the opposite sides be x meterThus, the third side = (54 – 2x) mArea (A) is given by;A = (54 – 2x)xA = 54x - 2x²We will differentiate to get;dA/dX = 54 - 4xFor a maximum Area we will differentiate and equate to zero i.e. dA/dx = 0. Thus;54 – 4x = 0 x = 54/4x = 13.5 mArea is 280 m². Thus, the other possible dimension is;w = 280/13.5w = 20.74 mPossible dimensions of the field are; 13.5 m by 20.74 m or 20.74 m by 13.5 mHow to maximize area?Since it is a rectangular plot, let the opposite sides be x meterThus, the third side = (54 – 2x) mArea (A) is given by;A = (54 – 2x)xA = 54x - 2x²We will differentiate to get;dA/dX = 54 - 4xFor a maximum Area we will differentiate and equate to zero i.e. dA/dx = 0. Thus;54 – 4x = 0 x = 54/4x = 13.5 mArea is 280 m². Thus, the other possible dimension is;w = 280/13.5w = 20.74 m