Answer:
[tex]300m\times 400m[/tex] or[tex]200m\times 600m[/tex]
Step-by-step explanation:
We are given that
Fencing used on three sides=1000 m
Area of field enclosure=120000 square meters
Let x be the length and y be the width of rectangle.
Fencing used on three sides=2x+y
[tex]2x+y=1000[/tex]
[tex]y=1000-2x[/tex]
Area of  field=[tex]xy=x(1000-2x)[/tex]
[tex]-2x^2+1000x=120000[/tex]
[tex]-x^2+500x=60000[/tex]
[tex]x^2-500x+60000=0[/tex]
Using quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{500\pm\sqrt{(500)^2-4(60000)}}{2}[/tex]
[tex]x=\frac{500\pm 100}{2}[/tex]
[tex]x=\frac{500+100}{2}=300m[/tex]
[tex]x=\frac{500-100}{2}=200 m[/tex]
[tex]y=1000-2(300)=400 m[/tex]
[tex]y=1000-2(200)=600m[/tex]
Dimension of the field
[tex]300m\times 400m[/tex] or[tex]200m\times 600m[/tex]
Answer:
length = 500 m , width = 250 m
Step-by-step explanation:
Let the length of the field is L and the width is W.
Area of the field , A = 120000 sq. metre
Length of fencing = 1000 m
Total length of fence = L + 2 W = 1000
L = 1000 - 2 W
Area = Length x width
A = L x W
A = (1000 - 2W) x W
A = 1000 W - 2 W²
for mxima and minima,
dA/dW = 0
1000 - 4W = 0
W = 250 m
L = 1000 - 2 x 250 = 500 m
Thus, the length of the field is 500 m and the width of the field is 250 m.
