A farmer has 200 m of fencing with which he wishes to enclose a rectangular field.One side of the field can make use of a fence that already exists. What is the maximum area he can enclose?

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A farmer has 200 m of fencing with which he wishes to enclose a rectangular field.One side of the field can make use of a fence that already exists. What is the maximum area he can enclose?

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Answer 1 · 2018-02-18T16:50:38

$5000 m^{2}$ $5000m^2$ is the required area. I have used elementary concepts of maxima and minima.

Explanation:

Let area be $A$ $A$ and the sides of rectangular field be $x and y;$ $x and y;$
So,
$A = x \cdot y$ $A=x*y$

Now, one side of the rectangle is already made with a fence.
There are 4 sides, two sides of $x$ $x$ meters and two sides of $y$ $y$ meters. Let a side of $y$ $y$ meters be already fenced.

Then, the remaining three sides are to be fenced with a fence of $200 m$ $200m$ length,

So,

$2 x + y = 200$ $2x+y=200$

$y = 200 - 2 x$ $y=200-2x$
So,

$A = x (200 - 2 x)$ $A=x(200-2x)$

$A = 200 x - 2 x^{2}$ $A=200x-2x^2$

For area to be maximum,

$\frac{d A}{d x} = 0$ $(dA)/dx=0$

And,

$\frac{d^{2} A}{{d x}^{2}} < 0$ $(d^2A)/dx^2<0$

Now,

$\frac{d A}{d x} = 200 - 4 x$ $(dA)/dx=200-4x$

$200 - 4 x = 0$ $200-4x=0$

$x = 50$ $x=50$

Now,

$\frac{d^{2} A}{{d x}^{2}} = - 4$ $(d^2A)/dx^2=-4$

Which is negative

Thus, $x = 50$ $x=50$ is a maximum.

Now,

$2 x + y = 200$ $2x+y=200$

$2 (50) + y = 200$ $2(50)+y=200$

$y = 100$ $y=100$

Thus, the required area is:

$A = x \cdot y = 50 \cdot 100 = 5000 m^{2}$ $A=x*y=50*100=5000 m^2$

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A farmer has 200 m of fencing with which he wishes to enclose a rectangular field.One side of the field can make use of a fence that already exists. What is the maximum area he can enclose?

1 Answer

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