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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:
https://brainly.com/question/19021959
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