9. A dog pen is to be built along a wall from 72 m . of fencing. Determine the value of $$ imes $$ that maximizes the area that will be enclosed.

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9. A dog pen is to be built along a wall from 72 m . of fencing. Determine the value of $$ 	imes $$ that maximizes the area that will be enclosed.

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We need to maximize the area of a rectangular dog pen that is built against a wall using 72 m of fencing. Let $$ x $$ be the length of the pen parallel to the wall, and $$ y $$ be the width of the pen perpendicular to the wall. The total fencing used will be $$ x + 2y = 72 $$.
The area $$ A $$ of the pen can be expressed as $$ A = x \cdot y $$. We can express $$ y $$ in terms of $$ x $$ using the fencing constraint.
From the equation $$ x + 2y = 72 $$, we can solve for $$ y $$:
$$
y = \frac{72 - x}{2}
$$
Now substitute this into the area formula:
$$
A = x \cdot \frac{72 - x}{2}
$$
Thus, the area function becomes:
$$
A(x) = \frac{72x - x^2}{2}
$$
This can be simplified to:
$$
A(x) = 36x - \frac{1}{2}x^2
$$
To find the value of $$ x $$ that maximizes the area, we need to take the derivative of $$ A(x) $$ and set it to zero:
$$
A'(x) = 36 - x
$$
Setting the derivative equal to zero gives:
$$
36 - x = 0
$$
Thus, $$ x = 36 $$.
The value of $$ x $$ that maximizes the area of the dog pen is $$ 36 $$ m.

A dog pen is to be built along a wall from 72 m . of fencing. Determine the value of that maximizes the
area that will be enclosed.

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A dog pen is to be built along a wall from 72 m . of fencing. Determine the value of that maximizes the area that will be enclosed.