Question

16. Peter is building a fenced in area for his dog. He has 250 feet (ft.) of fencing to create an area for his dog. He is building it next to his house so he will only need fencing for 3 sides of the rectangular area. The diagram shows the width, d, of the area Peter is building for his dog.

Answer 1

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.