Answer:
[tex]x=10[/tex] feet and [tex]y=20[/tex] feet
The least amount of fencing required = 40 feet.
Step-by-step explanation:
Let us take [tex]x[/tex] be the length of the garden and [tex]y[/tex] be width of the garden.
The area of the rectangular garden will be, [tex]A=x y=200[/tex] ...{1}
Perimeter of the garden, [tex]P=2 x+y[/tex].
[ because one side of the garden is protected by a barn ]
Thus,
[tex]P=2 x+y[/tex]
or
[tex]P=2 x+\frac{200}{x}[/tex] [using 1]
[tex]\frac{d P}{d x}=2-\frac{200}{x^{2}}[/tex]
For points of maxima or minima
[tex]\frac{d P}{d x}=0 \Rightarrow 2-\frac{200}{x^{2}}=0 \Rightarrow x^{2}=100 \Rightarrow x=\pm 10[/tex]
Since length cannot be negative, [tex]x=-10[/tex] is rejected.
Thus,
[tex]\frac{d^{2} P}{d x^{2}}=-(-2) \frac{200}{x^{3}}=\frac{400}{x^{3}}[/tex]
[tex]\left(\frac{d^{2} P}{d x^{2}}\right)_{x=10}=\frac{400}{10^{3}}>0[/tex]
Hence at [tex]x=10,[/tex], The perimeter [tex](P)[/tex] will be min.
On substituting [tex]x=10[/tex] in [tex]x y=200,[/tex] then [tex]y=20[/tex]
Hence, the dimensions of the garden are:
[tex]x=10[/tex] feet and [tex]y=20[/tex] feet
The perimeter, [tex]P=20+20=40[/tex] feet i.e the least amount of fencing required.
The lengths of the sides of the rectangular field so as to minimize the amount of fencing needed is 10 feet by 20 feet
Let x represent the length of the fence and y represent the width of the fence.
Since the area is 200, hence:
Area = length * width = x * y
200 = xy
y = 200/x
One side of the garden is already protected by a barn. Hence:
Amount of fencing needed (P) = x + x + y = 2x + y
Amount of fencing needed (P) = 2x + 200/x = 2x + 200/x
To minimize the amount of fencing needed, dP/dx = 0, hence:
dP/dx = 2 - 4050/x²
2 - 200/x² = 0
2 = 200/x²
2x² = 200
x² = 100
x = 10 feet
y = 200/x = 200/10 = 20 feet
Hence, the lengths of the sides of the rectangular field so as to minimize the amount of fencing needed is 10 feet by 20 feet
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