The amount of heavy-duty fencing to be used is 2000 ft for $9000 and the amount of standard fencing to be used is 3000 ft for $9000 in order to make a total cost of $18000.
The area of the rectangular plot is denoted by the length multiplied by the width.
Let us assume that the length = x and the width = y.
Then, for two opposite with heavy fencing for $4.50 a foot and the standard fencing for $3 a foot, we have:
2x (4.50) + 2y(3) = 18000 ---- (1)
⇒ 9x + 6y = 18000
Make (y) the subject, we have:
[tex]\mathbf{y = \dfrac{18000-9x}{6}}[/tex]
Replace the value of y with the area of the rectangle, and we get:
[tex]\mathbf{Area =x(\dfrac{18000 - 9x}{6})}[/tex]
Area = x(3000 - 1.5x)
Area = 3000x - 1.5x²
For the area to be maximum, we take the differentiation of the Area:
dA/dx = 0
dA/dx = 3000 - 3x
x = 3000/3
x = 1000 ft
From equation (1)
2x (4.50) + 2y(3) = 18000
x (4.50) + y(3) = 9000
1000(4.50) + 3y = 9000
3y = 9000 -4500
y = 4500/3
y = 1500 ft
So, there are (1000× 2)ft = 2000ft heavy duty fencing for $9000, and (1500 ×2)ft = 3000ft standard fencing for $9000 to make a cost of $18000.
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