Area of a rectanguar field with given fence length

Related BrainMass Solutions

• Length, Width, Perimeter and Area of a Rectangular Field

  Then, Area of the field = A = (LP/2) - L^2 = (150 * 600/2) - 150^2 = 22500 ft^2 [This will be the Maximum area the field can have with the given fence length.]

• Maximizing the Area of a Rectangle

  Then, Area of the field = A = (LP/2) - L^2 = (150 * 600/2) - 150^2 = 22500 ft^2 This will be the Maximum area the field can have with the given fence length.

• Minimization

  26452 Area A rancher wants to fence in an area of 1000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

• Minimization

  26448 Area and perimeter A rancher wants to fence in an area of 1000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

• Sample Question: Word problems

  A rancher wants to fence in an area of 2500000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use? 3.

• Derivatives : Maximum and Minimum Values

  Answer 50km/hr 8) A rectangular field , to enclose a given area , is fenced off along a straight river.

• Functions and models of cost

  319010 Functions and models of cost A wall is to be constructed around 3 sides of a rectangle of area 220 square yards, with a fence running the length of one side.

• minimization

  required for a given area.

• Area of a rectangle

  2456 Area of a rectangle A rectangular lot, 54 square yds. in area with a perimeter fence, is divided into 2 rectangular sections by a single connecting fence costing $2.00/yd. the perimeter fence costs $5.00/yd.