a farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. no fencing is needed along the barn, and the fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. if the fencing costs per linear foot to install and the farmer is not willing to spend more than , find the dimensions for the plot that would enclose the most area.

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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