SOLUTION: Please help! Submitted question earlier didn't get a response.
4) Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plu
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Question 36209: Please help! Submitted question earlier didn't get a response.
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.
Found 2 solutions by venugopalramana, Prithwis:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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:
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
Answer by Prithwis(166) (Show Source): You can put this solution on YOUR website!
Let the length of the rectangular patio be x ft
Perimeter of the rectangular patio = 400 ft
The width of the rectangular patio would be 1/2(400-2x) = 200-x
Area = Length * Width = x(200-x)
We need to maximize f(x) = x(200-x) to achieve the goal of the problem.
f(x) is quadratic function for Area, which represents a parabola opening downward (because the co-efficient of x^2 is negative).
Maximum of f(x) is reached at the vertex (because it opens downward)
x-Coordinate of the vertex = -b/2a (for ax^2+bx+c);
f(x) = -x^2+200x; So, a = -1, b=200, c=0
(-b/2a) = 100;
So, maximum area can be obtained if the length is 100 feet;
The width will be 1/2(400-200) = 100 feet;
Answer - Dimension of the patio for maximum area will be 100 feet X 100 feet
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