Respuesta :
Answer:
length = 200 m
width = 400 m
Step-by-step explanation:
Let the length of the plaing area is L and the width of the playing area is W.
Length of fencing around three sides = 2 L + W = 800
W = 800 - 2L ..... (1)
Let A is the area of playing area
A = L x W
A = L (800 - 2L)
A = 800 L - 2L²
Differentiate with respect to L.
dA/dL = 800 - 4 L
It is equal to zero for maxima and minima
800 - 4 L = 0
L = 200 m
W = 800 - 2 x 200 = 400 m
So, the area is maximum if the length is 200 m and the width is 400 m.
Answer:
Dimensions : Length = 400 , Width = 200
Step-by-step explanation:
Let length & width of rectangle park be = L , W . Also, let one side of length be supported by building wall, so not needing fencing.
So, the perimeter of rectangle park, including 2 width & 1 length :
2W + L = 800
L = 800 - 2W
Rectangle Area = Length x Width
A = L x W
A = (800 - 2W) W
A = 800W - 2W^2
To maximise area, it will have to be differentiated w.r.t dimension width
dA / dW = 800 - 4W
dA / dW = 800 - 4W = 0
800 = 4W
W = 800 / 4
Width [W] = 200
Length [L] = 800 - 2W
= 800 - 2(200)
= 800 - 400
Length [L] = 400
