Let's denote the width of the rectangular area as \( d \) and the length as \( l \). Since Peter is building a fenced area that borders the house, he only needs to fence 3 sides of the rectangle: two widths and one length.
The fencing needed can be expressed as:
\[ \text{Total fencing} = 2d + l \]
Peter has a total of 250 feet of fencing available, so we can set up the equation:
\[ 2d + l = 250 \]
To express the length \( l \) in terms of width \( d \), we can rearrange the equation:
\[ l = 250 - 2d \]
Now, the area \( A \) of the rectangular fenced area is given by the formula:
\[ A = l \times d \]
Substituting the expression for \( l \) we derived:
\[ A = (250 - 2d) \times d \]
Expanding this gives:
\[ A = 250d - 2d^2 \]
Thus, the area can be expressed as a quadratic function:
\[ A = -2d^2 + 250d \]
This quadratic function represents a parabola that opens downwards (since the coefficient of \( d^2 \) is negative), and we can find the maximum area by determining the vertex of the parabola. The formula for the \( d \) coordinate of the vertex in a quadratic equation of the form \( ax^2 + bx + c \) is given by:
\[ d = -\frac{b}{2a} \]
In our case, \( a = -2 \) and \( b = 250 \):
\[ d = -\frac{250}{2 \cdot -2} = \frac{250}{4} = 62.5 \]
Now, we can plug this value of \( d \) back into the equation for \( l \):
\[ l = 250 - 2(62.5) = 250 - 125 = 125 \]
Thus, the dimensions of the fenced area that will maximize the area are:
- Width \( d = 62.5 \) feet
- Length \( l = 125 \) feet
In conclusion, Peter should build a fenced area that is 62.5 feet wide and 125 feet long for optimal space for his dog.
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