farmer encloses rect. garden using barn as one side

vickie beauchamp

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A farmer decides to enclose a rectangular garden using the side of his barn as one side of the rectangle. What is the maximum area the farmer can enclose with 60sq ft of fence My answer is 225 sq ft is this right? I have the length being 15 ft and the width being 15 ft. Is this right?
 
Re: word problem

vickie beauchamp said:
A farmer decides to enclose a rectangular garden using the side of his barn as one side of the rectangle. What is the maximum area the farmer can enclose with 60sq ft of fence My answer is 225 sq ft is this right? I have the length being 15 ft and the width being 15 ft. Is this right?
No. Since barn covers one side, then you're only using 45 feet of the 60 feet: kapish?

Your "60sq ft of fence" should be simply "60 feet of fence".
 
A farmer decides to enclose a rectangular garden using the side of his barn as one side of the rectangle. What is the maximum area the farmer can enclose with 60sq ft of fence My answer is 225 sq ft is this right? I have the length being 15 ft and the width being 15 ft. Is this right?

Rectangle of maximum area with fixed perimeter.

Considering all rectangles with a given 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.

Considering all rectangles with a given 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.

Considering all rectangles with the same area, the square results in the smallest perimeter for a given area.


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
 
vickie beauchamp said:
A farmer decides to enclose a rectangular garden using the side of his barn as one side of the rectangle. What is the maximum area the farmer can enclose with 60sq ft of fence My answer is 225 sq ft is this right? I have the length being 15 ft and the width being 15 ft. Is this right?

I think everyone else missed the part "using the side of his barn as one side."
Let x = the length of the garden
and y = the width of the garden.

2x + y = 60

A = xy

y = 60 - 2x

A (x) = x(60 - 2x) = 60x - 2x^2

This is a parabola, so the maximum occurs at the vertex.
Do you know how to find the coordinates of the vertex?

The garden is NOT a square. That would be true if all 4 sides were fenced in.
 
A farmer decides to enclose a rectangular garden using the side of his barn as one side of the rectangle. What is the maximum area the farmer can enclose with 60sq ft of fence My answer is 225 sq ft is this right? I have the length being 15 ft and the width being 15 ft. Is this right?

I think everyone else missed the part "using the side of his barn as one side."

Considering all rectangles with a given perimeter, one side being another straight boundry (the barn), the 3 sided rectangle enclosing the greatest area has a length to width ratio of 2:1I hope this has been of some help to you.

TchrWill

TchrWill
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