if a man has 2500 ft. of fencing and wants to fence off a rectangle that borders a straight river(look below)?

Question

He needs no fence along the river. What is the largest area field he can enclose? USING CALCULUS

Answer 1

The area of a rectangle is just length times width:

A = L * W

Because we are not enclosing the entire rectangle, the perimeter equation is not the standard perimeter equation for a rectangle (2L + 2W). Let's say the river lies along the width, so the instead of twice the width, we end up with just the width:

P = 2L + W

Substituting because we know P=2500, and solving for W, we get:

2500 = 2L + W
W = 2500 - 2L

Substitute that into the area equation:

A = L * W
A = L * (2500 - 2L)
A = 2500L - 2L^2

To find the maximum take the derivative of area with respect to length, and set it equal to zero:

dA/dL = 0
d(2500L - 2L^2)/dL = 0
2500 - 4L = 0
L = 2500/4 = 625

Going back to the perimeter equation:

P = 2L + W
2500 = 2(625) + W
2500 = 1250 + W
W = 1250

So, the maximum area, is when W (measurement along the river) = 1250 and L = 625

A = W*L
A = 1250*625
A = 781,250 square feet

Answer 2

Start with an equation
Let W be the width of the area
Let L be the length of the area (opposite the river)
2W + L = 2500
L = 2500 - 2W

W * L = A
W * (2500 - 2W) = A
2500W - 2W^2 = A

We want A where it is a maximum, find the zero.

A' = 2500 + 4W
2500 = 4W
W = 625

L = 2500 - 2W = 2500 - 1250 = 1250

The largest enclosed are is then (1250)(625) = 781250 M^2

Answer 3

Area = LW
perimeter = 2W +L =2500
then L = 2500-2W
Area = (2500-2W)W =2500W-2W^2
da = 2500-4W
set the derivative equal to zero to maximize:

2500-4W= 0
2500=4W
W=625 ft
L= 2500-1250 = 1250

the field will be 1250 feet along the river and 625 ft wide.

Answer 4

So he wants to form a fence, such that 2a + b = 2500

Hence b = 2500 - 2a

Now area A = ab

A = a(2500-2a) = 2500a - 2a^2

dA/da = 2500 -4a

Set equal to zero to get a = 625 ft

Hence b = 1250 ft

The max area is ab = 781,250sq ft

Answer 5

Optimization. 781250 feet squared.