SOLUTION: Can you please help me find the dimensinos of a rectangle with the great area whose perimeter is 30 feet. I made an educated guess and made possible solution sets for the higher po

Question 204755: Can you please help me find the dimensinos of a rectangle with the great area whose perimeter is 30 feet. I made an educated guess and made possible solution sets for the higher possible number combinations. From that I saw that the solution should be between 7 and 8 ((7,8) and (8,7) solution sets give me the 30 foot perimeter and the highest area as whole numbers). Visualizing, I am thinking that there has to be a maximum of a parabola, and that, since this is a real object, the negative side of the axis would not apply. I just don't know what to put together to the equation I feel must exist if say, the question involved a perimeter of 150 feet, with a great many more possble solution sets. Does this tie into Geometry and the area of a square being the greatest perimeter of rectangles of identical perimeter or something?
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you are correct ___ a square is the rectangle with the largest area for a given perimeter

let L=length and W=width

2L + 2W = 30 ___ L = 15 - W

Area = L * W

substituting ___ A = (15 - W) * W = 15W - W^2

the max (squared term is negative) of the parabola lies on the axis of symmetry
___ W = -b / 2a = 15 / 2 = 7.5

substituting ___ L = 15 - 7.5 = 7.5
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