Question 210781: Your parents need to create a rectangular area for your dog to play in outside. They have 360 feet of wiring to distribute in the backyard. If you want your parents to create the maximum area for your dog to play in, what should the dimesions be?
Thank You
Answer by solver91311(24713)   (Show Source):
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The amount of wire used must be the perimeter of the rectangular area. One thing we know from the start is that maximization of the area demands that we use all of the wire. Hence, we know the perimeter of the area is the amount of wire available.
Next, we know that the perimeter of a rectangular figure is given by:
where  is the length of the rectangle and  is the width.
Further, we know that the area of a rectangle given the length, , and the width, , is given by:
Solving the perimeter function for in terms of and  where we consider to be a constant, results in:
Substituting this expression for into the formula for area gives us a polynomial function for Area,  , in terms of the independent variable :
 = \frac{Pw - 2w^2}{2})
or, in standard form  = ax^2 + bx + c\right)) :
 = -w^2 + \frac{P}{2}w)
Since the lead coefficient is < 0, we know that the graph of this function is a convex down parabola and therefore the maximum value of the function, which is the maximum area, is found at the vertex of the parabola.
Using the formula for the independent variable value at the vertex of  = ax^2 + bx + c) , namely  , the value of that maximizes ) is given by:
Hence, the width of the rectangle that has the maximum area for a given perimeter is one-fourth of the perimeter. That means that the two widths comprise half of the perimeter so the two lengths must comprise the other half of the perimeter, hence the length of the rectangle must also be one-fourth of the perimeter and equal to the width.
Therefore, the rectangle with the maximum area for a given perimeter is a square with sides one-fourth of the perimeter. Divide 360 by 4 to get your answer.
John
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