Question

a farmer has 500m of fencing to build a rectangular enclosure along a river.No fencing is needed along the riverbank.what would be the maximum area enclosed by the fence?what would be the equation?

Answer 1

Draw a rough sketch of the rectangle. Let left side be river, perpendiculars to river (a), and parallel to river (b). . Perimeter is 2a +b = 500,,,,,River side does not need fence . b= 500 -2a . Area of rectangle is A= l*w = a*b . making a quick table a,,,,,,,,,,,,,b,,,,,,,,,,,,,Area 0,,,,,,,,,,,225,,,,,,,,,,,,,,,,,,,,0 100,,,,,,,300,,,,,,,,,,,30,000 124,,,,,,,252,,,,,,,,,,,31,248 125,,,,,,,250,,,,,,,,,,,31,250,,,,,,,,,,max area 126,,,,,,,248,,,,,,,,,,,31,248 200,,,,,,,100,,,,,,,,,,,20,000 250,,,,,,,,,,,0,,,,,,,,,,,,,,,,,,,,0 . dim are a=125, b=25, for a max area of 31,250 m^2

Answer 2

a farmer has 500m of fencing to build a rectangular enclosure along a river.No fencing is needed along the riverbank.what would be the maximum area enclosed by the fence?what would be the equation? ------------------------ Area = length*width 2 sides are the width, one is length 500 = l + 2w --> l = 500-2w Area = l*w Sub for w A = w*(500-2w) A = 500w - 2w^2 ***** That's the equation ------------------ Method 1 Set the 1st derivate to zero 0 = 500 - 4w w = 125 l = 250 -------- Method 2: A = 500w - 2w^2 2w^2 - 500w + A = 0 This is a parabola. The max is at the vertex. The vertex is at w = -b/2a w = -b/2a = 500/4 w = 125 l = 250 same as above. ------------ Area = 125*250 = 31,250 sq meters.