Question

a farmer has 100m for fence avaibiable, with which he intends to build a pen for his sheep. he intends to create a rectangular pen against a permanent stone wall. i)show that a=1/2x(100-x) ii) find dA/dx and d2A/dx^2 iii)find the value of x that makes the area as large as possible, and explain how you know that this is a maximum.

Answer 1

The perimeter of the area of the pen the farmer intends to build for his

ship includes the length of the permanent stone wall.

Response:

i) The length and width of the rectangular pen are; x, and $\dfrac{100 - x}{2}$, therefore;

• The area is; $A = \dfrac{1}{2} \cdot x \cdot (100 - x)$

$ii) \hspace{0.5 cm}\dfrac{dA}{dx} = 50 - x$

$\dfrac{d^2A}{dx^2} = -1$

iii) The value of x that makes the area as large as possible is x = 50

How is the function for the area and the maximum area obtained?

Given:

The length of fencing the farmer has = 100 m

Part of the area of the pen is a permanent stone wall.

Let x represent the length of the stone wall, we have;

2 × Width = 100 m - x

Therefore;

Width, w, of the rectangular pen, $w = \mathbf{\dfrac{100 - x}{2}}$

Area of a rectangle = Length × Width

Area of the rectangular pen, is therefore;

• $A = x \times \dfrac{100 - x}{2} = \underline{\dfrac{1}{2} \cdot x \cdot (100 - x)}$

$ii) \hspace{0.5 cm} \mathbf{\dfrac{dA}{dx}}$, and $\mathbf{\dfrac{d^2A}{dx^2} }$ are found as follows;

$\dfrac{dA}{dx} = \mathbf{\dfrac{d}{dx} \left( \dfrac{1}{2} \cdot x \cdot (100 - x) \right)} = \underline{50 - x}$

$\dfrac{d^2A}{dx^2} = \mathbf{ \dfrac{d}{dx} \left( 50 - x\right)} = \underline{-1}$

iii) The value of x that makes the area as large as possible is given as follows;

Given that the second derivative, $\dfrac{d^2A}{dx^2} =-1$, is negative, we have;

At the maximum area, $\dfrac{dA}{dx} = \mathbf{0}$, which gives;

$\dfrac{dA}{dx} = 50 - x = 0$

x = 50

• The value of x that makes the area as large as possible is x  = 50

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