SOLUTION: The length and width of a rectangle must have a sum of 116 feet. find the dimensions of the rectangle whose area is as large as possible. answer should include length in feet and
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Question 608165: The length and width of a rectangle must have a sum of 116 feet. find the dimensions of the rectangle whose area is as large as possible. answer should include length in feet and width in feet.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is given by:
Which can be rearranged to:
The area, given by length times width, can then be represented as a function of the width for a given perimeter thus:
\ =\ \frac{Pw\ -\ 2w^2}{2})
This function is a quadratic that can be put into standard form thus:
\ =\ -w^2\ +\ \frac{P}{2}w)
The graph of such a quadratic is a parabola that is concave down (the lead coefficient is less than zero), meaning that the vertex represents a maximum. The 
coordinate of the vertex of this parabola is given by:
}\ = \frac{P}{4})
Therefore the width that gives the greatest area rectangle for any given perimeter is the perimeter divided by 4. The two widths are therefore half of the perimeter, hence the two lengths must also be half of the perimeter, and therefore the shape is a square.
Divide your given perimeter by 4 and then square the result.
John
My calculator said it, I believe it, that settles it
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