Respuesta :
Answer:
Therefore the dimension of the plot is 400 m × 200 m.
The area of the plot is = 80000 m²
Step-by-step explanation:
Given that a farmer ED has 800 meters of fencing, and wants to enclose a rectangular plot that borders on a river.
Let the length of side along river side be L.
And the width of the plot be be W.
Since the farmer does not fence the river side.
So the length of fence = (L+2w) m
According to the problem,
(L+2w) = 800
⇒L=800 - 2w
Therefore the area of the plot is A = Length ×width
= (L×W) m²
=[(800-2W)W] m²
For maximum or minimum [tex]\frac{dA}{dW}=0[/tex]
∴ A =(800-2W)W
⇒A = 800 W -2W²
Differentiating with respect to W
[tex]\frac{dA}{dW} = 800 -2.(2W)[/tex]
[tex]\Rightarrow \frac{dA}{dW} = 800 -4W[/tex]
For maximum or minimum [tex]\frac{dA}{dW}=0[/tex]
[tex]\Rightarrow 0 = 800 -4W[/tex]
[tex]\Rightarrow W= \frac{800}{4}[/tex]
⇒W=200
Again,
[tex]\frac{dA}{dW} = 800 -4W[/tex]
Again differentiating with respect to W
[tex]\frac{d^2A}{dW^2}= -4[/tex]
[tex]\therefore[ \frac{d^2A}{dW^2}]_{W=200}= -4<0[/tex]
When [tex]\frac{d^2A}{dW^2}<0[/tex] , then the area A is maximum at W=200.
When [tex]\frac{d^2A}{dW^2}>0[/tex] , then the area A is minimum at W=200.
Therefore at W= 200 m , the area of the plot maximum at W=200m
The length of the plot is L = [800- (2×200)] = 400 m
Therefore the dimension of the plot is 400 m × 200 m.
The area of the plot is = (400× 200) m²
= 80000 m²
The area is 80,000 m².
Fence all around rectangular area = 800 m
It means, this rectangular region's perimeter is: [tex]2l+w = 800\ ft[/tex]
since there is no fencing along the river's edge
Let the region's length (l) be x ft.
This region's breadth (w) is then [tex]= 800-2x.[/tex]
The area of the region:
[tex]\to A= l \times w = x(800-2x) = 800x-2x^2[/tex]
Differentiating
[tex]\to \frac{dA}{dx}=800-4x \\\\[/tex]
To maximize the area:
[tex]\to \frac{dA}{dx} = 0 \\\\ \to 800-4x = 0 \\\\\to 4x = 800 \\\\ \to x= \frac{800}{4}= 200\\\\[/tex]
Now the region's length is [tex]x= 200 \ m[/tex] and its breadth is[tex]800-2 (200) = 400[/tex] m. As a result, the rectangle region's maximum area:
[tex]\to \bold{A=200 \times 400 = 80,000 \ m^2}[/tex]
Learn more about the area:
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