Thinking problem

Question

A farmer has 300 metres of fence and wants to fence a rectangular plot of land adjacent to a straight river, using the river as one side of the rectangle so that only the remaining three sides (two short sides and one long side facing the river) need to be fenced.

(a) What dimensions of the plot will create the largest area that can be fenced?

(b) If the farmer wants to divide the plot into two equal areas by installing a fence parallel to the two short sides, what is the new optimal dimension?

By:Geometry Dash Lite

Answer 1

A farmer has 300 metres of fence and wants to fence a rectangular plot of land adjacent to a straight river, using the river as one side of the rectangle so that only the remaining three sides (two short sides and one long side facing the river) need to be fenced.

Given:

1. 300meters of fence
2. rectangular plot with one side a river , three sides a fence
3. 2 of the 3 sides are shorter than the long side that is parallel to river

Goal

1) find the dimensions of the plot will create the largest area that can be fenced?

River side

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Let x = the 2 sides of the fence

Let y = the side that is opposite the river

Equation 1: The length of the fencing is 300 meters.

// y is equal the the length of the fencing minus what was used for each side

The length of y then is y= 300m -2x

Equation 2: Area

The Area of the fenced in space is A = L x W

That would be Area = (x) (300-2x)

You can plot this as y=x(300-2x)

This is a parabola that opens down

The maximum area would be the vertex of the parabola.

The x value of the vertex is the midpoint of the zeros of the area equation

Area = x (300-2x) The zeros are when x=0

and when 300-2x = 0

300-2x +2x= +2x

300/2 = 2x / 2

150 = x

So the two zeros of the parabola are x=0 and x=150

The vertex is the midpoint of those values . Vertex is (150+0)/2 = 75

So when x =75, your area is maxed.

Plug 75 back into original equation and it will tell you the maximum area.

Area = x(300-2x)

Area - (75)(300-2(75))= 11250 meters squared

The optimum dimensions are

the 2 sides are x = 75 meters

the 3rd side is 300-2(75)= 150meters

Goal 2

(b) If the farmer wants to divide the plot into two equal areas by installing a fence parallel to the two short sides, what is the new optimal dimension?

Now you have 3 fence lengths that are x

y = 300-3x

Area = x(300-3x)

Your new zeros are x=0

and

300-3x=0

300=3x

100=x

The vertex of the parabola would be in the middle of 0 and 100.

So when x =50 the parabola would be maxed, and therefor area would be maxed

Plug x=50 into area equation to see maximum area

Area = x(300-3x)

Area = 50(300-3(50)) = 7500 Meters Squared.

The dimensions would be

The three sides would be x=50 meters

The side parallel to the river would be 300-3(50) = 150 meters