SOLUTION: can someone help me,i dont get this!!! Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet. i dont get this!!!

Question 92075This question is from textbook algebra and trigonometry
: can someone help me,i dont get this!!! Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet. i dont get this!!! This question is from textbook algebra and trigonometry

Found 2 solutions by mathispowerful, scott8148:
Answer by mathispowerful(115) (Show Source): You can put this solution on YOUR website!

First there are more than one way to do it, since this is algebra.com, I will use quadratic equation to solve it.
Let x, y represent length and width of the rectangle.
then we have x+y = 15
we need to find maximum area, A. A = length times width = xy
from x+y = 15, y = 15 - x
in A=xy, substitute y by 15 - x
A = x(15 - x)

this is a quadratic equation, the maximum value is at
so x = 7.5, then y = 15 -x = 7.5
so it will be a square with length of 7.5 ft.

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the perimeter (2L+2W) is 30 ... 2(L+W)=30 ... L+W=15 ... L=15-W ... so the area is A=(L)(W) ... A=(15-W)(W) ... A=15W-W^2

the maximum value for A occurs on the axis of symmetry of the graph

the equation of the axis of symmetry for ax^2+bx+c is x=-b/2a

in this case W=-15/2(-1) or W=7.5 ... since L+W=15, L=7.5

the rectangle of perimeter 30 with the greatest area is a square with side 7.5
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