This is an calculus problem:
Length + 2Width=336
Area= L*W = W*(336-2W)
Area= 336w-2W^2
Now on a graphing calculator, you can plot Area vs W to see where the maximum is.
In calculus, we take the derivative of Area, set it to zero, and solve for w.
dA/dw= 0=336-2*2W
W= 336/4=84ft
L= 336-2*84=168
max area=14112 sq ft.
I assume you are not in calculus yet. So graph it, and look for the max.
A developer wants to enclose a retangular grassy lot that borders a city street for parking. If the developer has 336 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed.
I went back and worked the problem again to make sure that it was right, and I ended up getting 84 X 84 which would be 7056 ft^2? But when I posted last time, I explained that the answer was 14,112 ft^2? Now I am not too sure of the answer.
5 answers
W= 336/4=84ft where did you get the 4 you divided by?
The derivative of 2W2 = 4W.
A developer wants to enclose a retangular grassy lot that borders a city street for parking. If the developer has 336 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed.
Considering all rectangles with a given perimeter, one side being another straight boundry, the 3 sided
rectangle enclosing the greatest area has a length to width ratio of 2:1.
\
Proof:
Let the width(the sides perpendicular to the given boundry) of the parcel be x.
The length of the parcel is then P - 2x.
The area of the parcel is A = x(P - 2x) = Px - 2x^2.
From the first derivitive, dA/dx = P - 4x making x = P/4 = 336/4 = 84 and P - 2x = 336 - 168 = 168.
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 168 with an area of 14,112 sq.ft.
Considering all rectangles with a given perimeter, one side being another straight boundry, the 3 sided
rectangle enclosing the greatest area has a length to width ratio of 2:1.
\
Proof:
Let the width(the sides perpendicular to the given boundry) of the parcel be x.
The length of the parcel is then P - 2x.
The area of the parcel is A = x(P - 2x) = Px - 2x^2.
From the first derivitive, dA/dx = P - 4x making x = P/4 = 336/4 = 84 and P - 2x = 336 - 168 = 168.
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 168 with an area of 14,112 sq.ft.
What is the largest area that can be enclosed? This is not just a calculus problem it's a zoning problem. If the city's Zoning Board doesn't approve the type of fencing the developer wishes to use, then the largest area that can be enclosed is exactly 0 feet.
1 answer