Question 37614: 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:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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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