straight river question

Question

A farmer has 100 meters of fencing and wants to enclose a rectangular field next to a straight river, using the river as one side of the enclosure. Since the river provides a natural boundary, fencing is only needed for the other three sides (two shorter sides and one longer side parallel to the river). What dimensions will maximize the enclosed area? Additionally, what is the maximum area that can be enclosed?

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Answer 1

A farmer has 100 meters of fencing and wants to enclose a rectangular field next to a straight river, using the river as one side of the enclosure. Since the river provides a natural boundary, fencing is only needed for the other three sides (two shorter sides and one longer side parallel to the river). What dimensions will maximize the enclosed area? Additionally, what is the maximum area that can be enclosed?

To find the dimension that will maximize the enclosed area with 100 meters of fencing next to a straight river, we can follow these steps:

Step 1: Define the Variables

Representation of x and y:

x = length of the side parallel to the river (the longer side).

y = length of each of he two sides perpendicular to the river (the shorter sides).

Step 2: Set Up the Perimeter Equation

Since the river serves as one side of the rectangle, we only need o enclose he other three sides. The total length of fencing used can therefore be expressed as:

x + 2y = 100

Step 3: Rearanging the Perimeter Equation

We can express y in terms of x:

Solve for y:

x + 2y = 100 Subtract (x) on both sides

2y = 100 - x Divide 2 on both sides

y = 100 - x divided by 2

Step 4: Use the Area Formula

Area of rectangle = Length x Width

length = x

width = y

A = x multiply by y

Substituting y = (100 - x divided by 2) in the area equation

Area = x times (100 - x divided by 2)

Step 5: Simplifying the Area Function

Area = x times (100 - x divided by 2)

A = 100x - x^2 (divided by 2)

A = 50x - x^2 (divided by 2)

Step 6: Find the Maximum Area

Rewrite the equation 50 -x = 0

Add (x) to both sides

50 = x

Step 7: Finding the value of y

Now, substitute x = 50 in the equation

y = 100 - x divided by 2

y = 100 -50 divided by 2

y = 50 divided by 2

y = 25

Step 8: Calculate the Maximum Area

Now, substitute x and y back into the area formula to find the maximum area:

A = x times y

A = 50 times 25

A = 1250

The dimensions that will maximize the enclosed area:

Length parallel to the river (x): 50 meters

Length perpendicular to the river (y): 25 meters

50m x 25m = 1250 square meters or 1250m^2

Final Solution:

The maximum area that can be enclosed is:

1250 square meters

I hope the mathematic calculations (step by step approach) is helpful. Please contact me anytime for

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Answer 2

Let W represent the length of the two shorter sides and L represent the length of the longer side parallel to the river. Since he has 100 feet of fencing the constraint on L and W is given by 2W + L = 100. This allows us to solve for L in terms of W. L = 100 - 2W.

The area of the rectangular field is A = LW. We can find a formula for the area of the field in terms of one variable by substituting for L.

A = (100 - 2W) W

Or A = 100W - 2W²

This is the equation of a downward opening parabola, so finding its vertex will determine the value of W that gives the maximum area.

There are two ways to find the vertex.

The axis of symmetry is given by W = -b/2a = -100/-4 = 25. When W = 25, L = 100 - 2(25) = 50.

So the dimensions that give the max area are W = 25, L = 50. and that max area is 25(50) = 1250.

Another way to determine the value for W that gives the max is to realize that the roots of the quadratic equation are 0 and 50. The axis of symmetry is the midpoint of the segment joining the roots (or the average of the roots: (0 + 50)/2 = 25.

Try graphing the function on Desmos as A(x) = -2x² + 100x and confirm that the vertex is located at (25, 1250).

Check your result here:

desmos.com/calculator/stm3r6i956

Answer 3

50 feet by 25 feet

Area = A = wL = width x Length. Length is parallel to the river, width is perpendicular to the river

100 = 2w+L

L = 100-2w

A = w(100-2w) = 100w -2w^2

A'= 100 -4w=0

w = 100/4= 25 feet

L = 50 feet

max Area = 25x50 = 1,250 ft^2

or, without differential calculus optimization

rewrite the Area equation in vertex form a(x-h)^2 +k with vertex (h,k)

-2w^2 +100w

=-2(w^2 -50w +625) + 1250

= -2(w-25)^2 -1250 with vertex (h,k) = (25, 1250)= maximum point