Answer:
Dimensions are
Length = 42.4264 feet
Width = 42.4264 feet
Step-by-step explanation:
Let the length be L and width be B
Data provided in the question:
Area of barn = LB = 1800 square feet
or
L = [ 1800 ÷ B ] feet
Amount of fencing = Perimeter of the barn, P = 2( L + B)
P = [tex]2(\frac{1800}{B}+B)[/tex]
differentiating the P with respect to B
we get,
[tex]\frac{dP}{dB}=2[\frac{(-1)1800}{B^2}+1][/tex]
for point of maxima or minima
[tex]\frac{dP}{dB}=2[\frac{(-1)1800}{B^2}+1]=0[/tex]
or
[tex][\frac{(-1)1800}{B^2}]=-1[/tex]
or
B² = 1800
or
B = 42.4264 feet
on again differentiating
[tex]\frac{d^2P}{dB^2}=2[\frac{(+2)1800}{B^3}[/tex]
here,
positive value mean at B = 42.43 feet perimeter is least
Therefore,
L = 1800 ÷ 42.426
= 42.4264 feet
Hence,
Dimensions are
Length = 42.4264 feet
Width = 42.4264 feet
The dimensions that will require the least amount of fencing is 30 feet by 60 feet
Let the dimensions of the garden be x and y.
Because one side of the fence is already protected, the perimeter of the garden would be:
[tex]\mathbf{P = 2x + y}[/tex]
And the area is:
[tex]\mathbf{A = xy}[/tex]
Substitute 1800 for A
[tex]\mathbf{xy = 1800}[/tex]
Make y the subject
[tex]\mathbf{y = \frac{1800}x}[/tex]
Substitute [tex]\mathbf{y = \frac{1800}x}[/tex] in [tex]\mathbf{P = 2x + y}[/tex]
[tex]\mathbf{P = 2x + \frac{1800}{x}}[/tex]
Differentiate
[tex]\mathbf{P' = 2 - \frac{1800}{x^2}}[/tex]
Set to 0
[tex]\mathbf{2 - \frac{1800}{x^2} = 0}[/tex]
Collect like terms
[tex]\mathbf{2 = \frac{1800}{x^2}}[/tex]
Multiply both sides by x^2, and divide both sides by 2
[tex]\mathbf{x^2 = 900}[/tex]
Take positive square roots of both sides
[tex]\mathbf{x = 30}[/tex]
Substitute 30 for x in [tex]\mathbf{y = \frac{1800}x}[/tex]
[tex]\mathbf{y = \frac{1800}{30}}[/tex]
[tex]\mathbf{y = 60}[/tex]
Hence, the dimensions that will require the least amount of fencing is 30 feet by 60 feet
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