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Answer and step-by-step explanation:
For any rectangle, the one with the largest area will be the one whose dimensions are as close to a square as possible.
However, the dividers change the process to find this maximum somewhat.
Letting x represent two sides of the rectangle and the 3 parallel dividers, we have 2x+3x = 5x.
Letting y represent the other two sides of the rectangle, we have 2y.
We know that 2y + 5x = 750.
Solving for y, we first subtract 5x from each side:
2y + 5x - 5x = 750 - 5x
2y = -5x + 750
Next we divide both sides by 2:
2y/2 = -5x/2 + 750/2
y = -2.5x + 375
We know that the area of a rectangle is given by
A = lw, where l is the length and w is the width. In this rectangle, one dimension is x and the other is y, making the area
A = xy
Substituting the expression for y we just found above, we have
A = x(-2.5x+375)
A = -2.5x² + 375x
This is a quadratic equation, with values a = -2.5, b = 375 and c = 0.
To find the maximum, we will find the vertex. First we find the axis of symmetry, using the equation
x = -b/2a
x = -375/2(-2.5) = -375/-5 = 75
Substituting this back in place of every x in our area equation, we have
A = -2.5x² + 375x
A = -2.5(75)² + 375(75) = -2.5(5625) + 28125 = -14062.5 + 28125 = 14062.5
The area of the four pen can be simplified to fi=nd the largest area
possible.
The largest possible total area of the four pens is 14,062.5 ft.²
(a) Diagrams illustrating the situation can be formed adjusting the width of
the included diagram illustrating the general situation.
(b) Please find the diagram illustrating the general situation created using
MS Visio.
(c) The expression for the total area is [tex]A = \left( 150 - \dfrac{2 \cdot Y}{5} \right) \cdot Y[/tex]
Reasons:
The given parameter are;
Length of fencing available = 750 ft.
Area to be enclosed = Rectangular
Number of included pens with parallel fencing = Four
Let Y represent the width of the area and let X represent the breadth
Therefore, we have;
2·Y + 5·X = 750
Which gives;
[tex]X = \dfrac{750}{5} - \dfrac{2 \cdot Y}{5} = \mathbf{150 - \dfrac{2 \cdot Y}{5}}[/tex]
The area is therefore;
[tex]A = Y \times \left( 150 - \dfrac{2 \cdot Y}{5}\right) = 150 \cdot Y - \dfrac{2 \cdot Y^2}{5}[/tex]
- [tex]\underline{\mathrm{Total \ area, \ A} = 150 \cdot Y - \dfrac{2 \cdot Y^2}{5}}[/tex]
[tex]\dfrac{dA}{dY} = \dfrac{d}{dY} \left ( 150 \cdot Y - \dfrac{2 \cdot Y^2}{5} \right) = 150 - \dfrac{4 \cdot Y}{5} = 0[/tex]
- [tex]150 - \dfrac{4 \cdot Y}{5} = 0[/tex]
[tex]Y= \dfrac{150 \times 5}{4} = 187.5[/tex]
The width of the maximum area, Y = 187.5 ft.
Which gives;
[tex]X = 150 - \dfrac{2 \times 187.5}{5} = 75[/tex]
The breadth of the area, X = 75 ft.
The maximum area, [tex]A_{max}[/tex] = Y × X, which gives;
[tex]A_{max}[/tex] = 187.5 ft. × 75 ft. = 14,062.5 ft.²
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