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 tp maxamize the area of his patio (area of a rectangle is length times width) What should the dimensions of the patio be,and show how themaxium area of the patio is calculated from the alderbraic equation.
2006-11-24 12:20:54 · 6 answers · asked by kiwikofii 2 in Science & Mathematics ➔ Mathematics
2l + 2w = 300
2l = 300 - 2w
l = 150 - w
Maximize length times width:
A(w) = w(150 - w) = 150w - w²
To maximize, take derivative, set = 0, and solve for w.
150 - 2w = 0
2w = 150
w = 75
and then l = 150 - 75 = 75.
2006-11-24 12:25:42 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
Here it goes,
Perimeter of the rectangle is 300 feet by your statement of the problem.
P=2L+2W
To maximize the area of this patio your rectangle would have to be a square. (A little bit of geometry would help you here but I will leave that out)
We know that area of a rectangle is equal to A=L*W
So let's solve P=2L+2W for either W or L, I chose L
L=150-W
So the area equals,
A=(150-W)W
A=150W-W^2 (W squared)
If you were to graph this it would be a downward shaping parabola with a maximum width amount at the vertex (think w for x and A for y) So W=75 at the Maximum so We have a patio of dimension:
75ft x75ft. With a Maximum area of: 5625 ft. (75*75).
2006-11-24 12:32:45 · answer #2 · answered by andrew b 1 · 0⤊ 0⤋
The largest rectangle in area is a square, so the patio ought to be 75 X 75. Algebraically, I guess that would be 4x = 300.
2006-11-24 12:27:33 · answer #3 · answered by bgdddymtty 3 · 0⤊ 0⤋
The dimensions will be a 75 x 75 ft square. (Yes a square is a rectangle!)
This actually takes some elementary calculus to prove:
A = L*W and P = 300 = 2L + 2W---->W = 150 - L the area eq is then:
A = L*(150-L) = 150L - L²
dA/dL = 150 - 2L = 0 ----> L = 75 -----> W = 75
2006-11-24 12:32:03 · answer #4 · answered by Steve 7 · 0⤊ 0⤋
unfortunately you fail to specify LINEAR or RUNNING feet...this makes the problem seem unanswerable. Obviously 300 feet divided x 4= 75 feet this is the square version but since the question speaks of "area" and perimeter both its seems to me to be an invalid question...sorry I could be of more assistance.
2006-11-24 12:29:16 · answer #5 · answered by atomic49er 3 · 0⤊ 0⤋
no matter what combination length and width you use, the area will always be the same.. for an algebraic equation i don feel like typing the problem
2006-11-24 12:25:53 · answer #6 · answered by xreighted s 2 · 0⤊ 2⤋