SOLUTION: John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a

Question 51306: John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation.
Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex form to find the maximum area.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
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Amanda has 400 feet of lumber to frame a
rectangular patio (the perimeter of a rectangle is 2
times length plus 2 times width). She wants to
maximize the area of her patio (area of a rectangle is
length times width). What should the dimensions of the
patio be, and show how the maximum area of the patio
is calculated from the algebraic equation.
Answer:
IF L AND B ARE DIMENSIONS WE HAVE
PERIMETER=2(L+B)=400.....OR.....L+B=200..OR......B=200-L.................I
AREA=A
=LB=L(200-L)=200L-L^2=-{L^2-200L}=-{(L^2)-2(L)(100)+100^2-100^2}
A=10000-(L-100)^2
(L-100)^2 BEING PERFECT SQUARE,ITS MINIMUM VALUE IS
ZERO.
HENCE AREA IS MAXIMUM WHEN L-100 IS ZERO,OR WHEN L=100
AND THEN THE MAXIMUM AREA WOULD BE
A-MAX.=10000-0=10000
DIMENSIONS ARE 100*100

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