Respuesta :
The constraint for the maximum enclosed area is the amount the farmer is
willing to spend on fencing.
The width of the plot that will give the most area, y = 83.[tex]\underline{\overline 3}[/tex] feet
The length of the plot that will give the most area, y = 125 feet
Reasons:
The given parameters are;
The cost of fencing per foot = $20
The maximum amount the farmer is willing to spend = $5000
Number of sides of fencing = 3 sides
Cost of the west side of the = Split with neighbor
Solution:
Let, y, represent the length of the fence, and let x represent the width of
the fence, we have;
Length of fence required = y + 2·x
Cost of the fence = 20·y + 20·x + (20÷2)·x = 20·y + 30·x
Therefore;
Maximum amount to be spent on the fence, 5000 = 20·y + 30·x
[tex]\therefore y = \dfrac{5000 - 30 \cdot x}{20} = 250 - \dfrac{3}{2} \cdot x[/tex]
[tex]y = \mathbf{250 - \dfrac{3}{2} \cdot x}[/tex]
Area of a rectangle = Length × Width
The enclosed rectangular area of the fencing, A = y × x
[tex]\therefore A = \left(250 - \dfrac{3}{2} \cdot x\right) \cdot x = 250 \cdot x - \dfrac{3}{2} \cdot x^2[/tex]
[tex]\therefore A = -\mathbf{ \dfrac{3}{2} \cdot x^2 + 250 \cdot x}[/tex]
The leading coefficient of the quadratic function of the area is negative,
therefore, the function only has a maximum point.
At the point of the most area, the slope, [tex]\dfrac{dA}{dx} = 0[/tex]
Finding the value of x at the maximum point, we get;
[tex]\dfrac{dA}{dx} = \mathbf{ \dfrac{d}{dx} \left( 250 \cdot x - \dfrac{3}{2} \cdot x^2\right)} = 250- 3 \cdot x = 0[/tex]
Therefore;
[tex]\mathrm{ The \ width \ of \ the \ plot \ that \ will \ enclose \ the \ area}, \ x = \dfrac{250}{3} \ feet = \underline{83.\overline 3 \ feet}[/tex]
[tex]\mathrm{ The \ length \ of \ the \ plot \ that \ will \ enclose \ the \ area}, \ y = 250 - \dfrac{3}{2} \times \dfrac{250}{3} = 125[/tex]
The length of the plot that will give the most area, y = 125 feet
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