SOLUTION: Please help 4)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

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Question 35991: Please help

4)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:
Show work in this space.

Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY
LET LENGTH BE L AND WIDTH BE W
WE HAVE PERIMETER =2(L+W)=300
L+W=150
W=150-L.............................I
AREA =LW=L(150-L)= - (L^2-150L) = - {(L^2-2*L*75+75^2)-75^2}
= 75^2 - (L-75)^2.......SINCE (L-75)^2 IS ALWAYS POSITIVE ,AND IT IS
TO BE SUBTRACTED FROM 75^2 TO GET THE AREA,WE GET MAXIMUM AREA WHEN
THIS IS MINIMUM...OR...L-75=0....OR....L=75
SO FOR MAXIMUM AREA THE PATIO SHOULD BE A SQUARE OF 75' LENGTH AND 75' WIDTH.

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