The rectangular field have four sides, where the opposite sides of the field are equal
The length of the brick wall that gives the lowest total cost of the fence is 40 meters
Let the length of the rectangular field be x, and the width be y.
Where: y represents the side to be made of brick wall,
So, the perimeter of the field is calculated using:
[tex]\mathbf{P =2x + 2y}[/tex]
And the area is
[tex]\mathbf{A =xy}[/tex]
The area is given as 2400.
So, we have:
[tex]\mathbf{xy = 2400}[/tex]
Make x the subject in
[tex]\mathbf{x = \frac{2400}y}[/tex]
Rewrite the perimeter as:
[tex]\mathbf{P =2x + y + y}[/tex]
The brick wall is $10 per meter, while the wooden wall is $20 per 4 meters
So, the cost function becomes
[tex]\mathbf{C =\frac {20}4 \times (2x + y) + 10 \times y}[/tex]
[tex]\mathbf{C =5 \times (2x + y) + 10 \times y}[/tex]
Open brackets
[tex]\mathbf{C =10x + 5y + 10y}[/tex]
[tex]\mathbf{C =10x +15y}[/tex]
Substitute [tex]\mathbf{x = \frac{2400}y}[/tex] in the cost function
[tex]\mathbf{C =10 \times \frac{2400}{y} +15y}[/tex]
[tex]\mathbf{C = \frac{24000}{y} +15y}[/tex]
Differentiate
[tex]\mathbf{C' = -\frac{24000}{y^2} +15}[/tex]
Set to 0, to minimize
[tex]\mathbf{-\frac{24000}{y^2} +15 = 0}[/tex]
Rewrite as
[tex]\mathbf{\frac{24000}{y^2} =15}[/tex]
Divide through by 15
[tex]\mathbf{\frac{1600}{y^2} =1}[/tex]
Multiply both sides by y^2
[tex]\mathbf{y^2 =1600}[/tex]
Take square roots of both sides
[tex]\mathbf{y^2 =40}[/tex]
Hence, the length of the brick wall should be 40 meters
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