True. It will use up valuable space.
find out the total wight and then each individual weight and then you can subtract
Answer:
the answer is border setting
If the rectangular field has notional sides
x
and
y
, then it has area:
A
(
x
)
=
x
y
[
=
6
⋅
10
6
sq ft
]
The length of fencing required, if
x
is the letter that was arbitrarily assigned to the side to which the dividing fence runs parallel, is:
L
(
x
)
=
3
x
+
2
y
It matters not that the farmer wishes to divide the area into 2 exact smaller areas.
Assuming the cost of the fencing is proportional to the length of fencing required, then:
C
(
x
)
=
α
L
(
x
)
To optimise cost, using the Lagrange Multiplier
λ
, with the area constraint :
∇
C
(
x
)
=
λ
∇
A
∇
L
(
x
)
=
μ
∇
A
⇒
μ
=
3
y
=
2
x
⇒
x
=
2
3
y
⇒
x
y
=
{
2
3
y
2
6
⋅
10
6
sq ft
∴
{
y
=
3
⋅
10
3
ft
x
=
2
⋅
10
3
ft
So the farmer minimises the cost by fencing-off in the ratio 2:3, either-way
