a. The dimensions that should be used to make the gardens are length = 25 yards and breadth = 33.33 yards
b. The largest area possible will be 3333.33 yards²
Since the Two rectangular gardens are made from 200 yards of fencing, let
Since they have a common side which is the breadth of the rectangular area, the total length of fencing L' = 4L + 3B
Since L' = 200 yards,
4L + 3B = 200
Making L subject of the formula, we have
L = (200 - 3B)/4
Since the area of the two rectangular gardens is A = 2L × 2B = 4LB
Substituting L into A, we have
A = 4LB
= 4(200 - 3B)/4 × B
= 200B - 3B²
The dimensions that should be used to make the gardens are length = 25 yards and breadth = 33.33 yards
To find the value of B at which A is maximum, we differentiate A with respect to B and equate to zero.
So, dA/dB = d(200B - 3B²)/dt
= 200 - 6B
Equating to zero, we have
dA/dB = 0
200 - 6B = 0
200 = 6B
B = 200/6
B = 33.33 yards
We need to determine if this value of B maximizes A. So, we differentiate A twice.
So, d²A/dB² = d(200 - 6B)/dB = -6 < 0. So, A is maximum at B = 200/6
Since L = (200 - 3B)/4
Substituting the value of B into the equation, we have
L = (200 - 3B)/4
L = (200 - 3 × 200/6)/4
L = (200 - 100)/4
L = 100/4
L = 25 yards
The dimensions that should be used to make the gardens are length = 25 yards and breadth = 33.33 yards
The largest area possible will be 3333.33 yards²
Since A = 4LB
Substituting the values of the variables into the equation, we have
A = 4LB
A = 4 × 25 × 200/6
A = 20000/6
A = 3333.33 yards²
So, the largest area possible will be 3333.33 yards²
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