Your answer to the first problem is correct.
Let x be the dimension parallel to the street. Since the sum of three sides must be 336, the other two side lengths are both (336 -x)/2 = 168 - x/2
The area enclosed by the three sides is
A = (168 - x/2)*x = 168 x - x^2/2
The derivative, which must be zero at maximum area, is
dA/dx = 168 - x = 0
Therefore x = 168 and the maximum area is
A = 168*84 = 14,112 ft^2
I have a couple of questions. Could you please check one of them and help me with one?
1. You have a 256 feet of fencing to enclose a retangular region. Find the demensions of the rectangle that maximize the enclosed area?
This is what I did:
a=xy
2x+2y=236
I subtacted 2x from each side
2y=236-2x
I divided by 2
Now I have: y=118-x
a=xy=x(118-x)
a(x)=x(118-x)
f(x)=ax^2+bx+c
a(x)=x(118-x)=118x-x^2
-x^2+118x
x=b/2a=-118/2(-1)=59
2(59)+2y=236
118+2y=236
2y=118
I divided by 2 on each side
y=59
Fianl answer: 59 ft by 59 ft
2.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.
answer choices: a. 14,112 ft^2, b.28,224 ft^2, c.21,168 ft^2, d. 7056 ft^2
-This question I don't understand or how to start
4 answers
1. You have a 256 feet of fencing to enclose a retangular region. Find the demensions of the rectangle that maximize the enclosed area?
Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
Therefore, (256/4)^2 = a square 64 by 64, the rectangle enclosing the maximum area of 4096 sq.ft. with a perimeter of 256 ft.
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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
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 128 with an area of 14,112 sq.ft.
I'll let you prove this to yourself.
Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
Therefore, (256/4)^2 = a square 64 by 64, the rectangle enclosing the maximum area of 4096 sq.ft. with a perimeter of 256 ft.
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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
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 128 with an area of 14,112 sq.ft.
I'll let you prove this to yourself.
Correction of typo:
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
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 168 with an area of 14,112 sq.ft. (168 was incorrectly typed as 128)
I'll let you prove this to yourself.
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
Therefore, the maximum size rectangle with a side ratio of 2:1 is 84 by 168 with an area of 14,112 sq.ft. (168 was incorrectly typed as 128)
I'll let you prove this to yourself.
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.