largest area that can be enclosed is BLANK square meters.

Question

farmer ed has 8,500 meters of fencing, and wants to enclose a rectangular plot that borders on a river. if farmer ed does not fence the side along the river, what is the largest area that can be enclosed?

Answer 1

Let's say the length of sides perpendicular to the river = x

and, the length of side parallel to the river = y

 

Since the side along the river is not fenced, only 3 sides remain to be fenced.
2x + y = 8500  eq1

 

Area of the rectangle = x*y

 

substitute value of y from eq1

area = x(8500-2x)

=-2x² + 8500x        eq2

 

Since the coefficient of x2 is negative, this is a parabola opening down,

in which case it will have it's maximum value at the vertex.

 

Now, the vertex of parabola with standard form f(x) = ax² + bx + c is at x=-b/2a

so, the vertex of parabola represented by eq2 will be at x = -8500/2(-2) = 2125

This means that area will be maximum when x = 2125

 

We plug this value in eq2 to get the maximum area.

Maximum area = -2(2125)² + 8500(2125) = 9,031,250 meters²

 

 

Answer 2

2L + B = 8500

A=LB = L(8500-2L)= 8500L-2L²

dA/dL = 8500 - 4L = 0

L = 2125 m

B = 8500 - 4250 = 4250 m

A=2125x4250=9031250 m²

Answer 3

This is a classic optimization problem.

 

The formula for the area of a rectangle is  A = L W   (L= length,  W = width)

 

We would like to optimize A, but A depends on two variables not just one.

 

To fix this problem, we use the equation 8500 = L + 2W.     This is called the constraint equation.   It comes about because the farmer must fence the width twice, but the length only once (the river is the other "fence").

 

The constraint equation can be reorganized as   L =  8500 - 2W.   Substituting this into the areas equation gives

 

  A = (8500 - 2W) W =  8500 W - 2W²  .

 

This now depends on just one variable (W), so we can find the optimal value of W by differentiating this expression  for A  with respect to W and then setting the resulting derivative equal to zero.

 

    A'  =  8500 - 4W   = 0         This implies that the optimal value of width is  W =  8500 /4.

 

  Substituting back into   L =  8500 - 2W     implies   that the optimal value of length is L =  8500/2.

 

 

Notice that L  turns out to be twice W. 

 

 

The optimal area  is just  A =  L W =  (8500)² /8

 

Answer 4

Let y = length of the side parallel to the river

       x = length of each of the two sides perpendicular to the river

 

Then, y + 2x = 8500.  So, y = 8500 - 2x

 

Area = xy = x(8500 - 2x) = -2x² + 8500x

 

The graph of the area function is a parabola opening downward, so it has a maximum.  

 

The maximum occurs when x = x-coordinate of the vertex

 

                                                    = -8500/(2(-2)) = 8500/4 = 2125

 

Maximum area = -2(2125)² + 8500(2125)  = 9,031,250 m²