Question

A farmer wants to build a rectangular pen which will be bounded on one side by a river and on the other three sides by a single-strand electric fence. If the farmer has 40 meters of wire to use, what is the largest area that the farmer can enclose?

Answer 1

Answer:

A) Dimensions;

Length = 20 m and width = 10 m

B) A_max = 200 m²

Step-by-step explanation:

Let x and y represent width and length respectively.

He has 40 metres to use and he wants to enclose 3 sides.

Thus;

2x + y = 40 - - - - (eq 1)

Area of a rectangle = length x width

Thus;

A = xy - - - (eq 2)

From equation 1;

Y = 40 - 2x

Plugging this for y in eq 2;

A = x(40 - 2x)

A = 40x - 2x²

The parabola opens downwards and so the x-value of the maximum point is;

x = -b/2a

Thus;

x = -40/2(-2)

x = 10 m

Put 10 for x in eq 1 to get;

2(10) + y = 40

20 + y = 40

y = 40 - 20

y = 20m

Thus, maximum area is;

A_max = 10 × 20

A_max = 200 m²

Answer 2

The area of the pen is the product of its dimensions

The largest area the farmer can enclose is 200 square meters

The perimeter of the pen is given as:

$\mathbf{P = 40}$

Because one side of the fence is covered by a river, the perimeter would be:

$\mathbf{2x + y = 40}$

Make y the subject

$\mathbf{y = 40 - 2x}$

The area of the pen is:

$\mathbf{A = xy}$

Substitute $\mathbf{y = 40 - 2x}$

$\mathbf{A = x(40 - 2x)}$

Expand

$\mathbf{A = 40x - 2x^2}$

Differentiate

$\mathbf{A' = 40 - 4x}$

Set to 0

$\mathbf{ 40 - 4x = 0}$

Rewrite as:

$\mathbf{ 4x = 40}$

Divide through by 4

$\mathbf{x = 10}$

Substitute 10 for x in $\mathbf{A = x(40 - 2x)}$

$\mathbf{A = 10 \times (40 - 2(10))}$

$\mathbf{A = 200}$

Hence, the largest area the farmer can enclose is 200 square meters

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