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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.

Sagot :

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 [tex]\dfrac{100 - x}{2}[/tex], therefore;

  • The area is; [tex]A = \dfrac{1}{2} \cdot x \cdot (100 - x)[/tex]

[tex]ii) \hspace{0.5 cm}\dfrac{dA}{dx} = 50 - x[/tex]

[tex]\dfrac{d^2A}{dx^2} = -1[/tex]

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, [tex]w = \mathbf{\dfrac{100 - x}{2}}[/tex]

Area of a rectangle = Length × Width

Area of the rectangular pen, is therefore;

  • [tex]A = x \times \dfrac{100 - x}{2} = \underline{\dfrac{1}{2} \cdot x \cdot (100 - x)}[/tex]

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

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

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

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

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

At the maximum area, [tex]\dfrac{dA}{dx} = \mathbf{0}[/tex], which gives;

[tex]\dfrac{dA}{dx} = 50 - x = 0[/tex]

x = 50

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

Learn more about the maximum value of a function here:

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