MATH SOLVE

2 months ago

Q:
# A rancher has 140 feet of fence with which to enclose three sides of a rectangular garden (the fourth side is a cliff wall and will not require fencing). Find the dimensions of the garden with the largest possible area. (For the purpose of this problem, the width will be the smaller dimension (needing two sides); the length with be the longer dimension (needing one side).)

Accepted Solution

A:

A rancher wants to enclose a rectangular field with 220 ft of fencing.

One side is a river and will not require a fence.

What is the maximum area that can be enclosed?

:

The field requires only 3 sides of fence, therefore

L + 2W = 220

L = (220-2W); we can use this for substitution in the Area equation

:

Area

A = L*W

replace L with (220-2W)

A = W(220-2W)

A = -2W^2 + 220W

A quadratic equation, max A occurs at the axis of symmetry, x = -b/(2a)

In this equation

W = %28-220%29%2F%282%2A-2%29

W = 55 ft is the width for max area

:

Find the max area

A = -2(55^2) + 220(55)

A = -2(3025) + 12100

A = 6050 sq/ft is the max area

:

:

Confirm this with the dimensions calculated; 110 * 55 = 6050

One side is a river and will not require a fence.

What is the maximum area that can be enclosed?

:

The field requires only 3 sides of fence, therefore

L + 2W = 220

L = (220-2W); we can use this for substitution in the Area equation

:

Area

A = L*W

replace L with (220-2W)

A = W(220-2W)

A = -2W^2 + 220W

A quadratic equation, max A occurs at the axis of symmetry, x = -b/(2a)

In this equation

W = %28-220%29%2F%282%2A-2%29

W = 55 ft is the width for max area

:

Find the max area

A = -2(55^2) + 220(55)

A = -2(3025) + 12100

A = 6050 sq/ft is the max area

:

:

Confirm this with the dimensions calculated; 110 * 55 = 6050