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High School
Math Word Problem

nasriibraahim507

Answered question

2022-04-07

A farmer has a 200 m fencingmaterial to enclose two adjacent sides of a rectangular field. What dimensions should be used so that the enclose area will be a maximum?

Answer & Explanation

nick1337

Expert2022-04-19Added 777 answers

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

$A = x \cdot y$

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

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

So,

$2 x + y = 200$

$y = 200 - 2 x$
So,

$A = x (200 - 2 x)$

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

For area to be maximum,

$\frac{d A}{d x} = 0$ and $\frac{d^{2} A}{{d x}^{2}} < 0$

Now,

$\frac{d A}{d x} = 200 - 4 x$

$200 - 4 x = 0$

$x = 50$

Now,

$\frac{d^{2} A}{{d x}^{2}} = - 4$

Which is negative

Thus, $x = 50$ is a maximum.

Now,

$2 x + y = 200$

$2 (50) + y = 200$

$y = 100$

Thus, the required area is:

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

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