Extreme Values: fencing grazing area next to river

[Idealistic]

A farmer wants to put a rectangular perimeter around some land that just so happens to by near a river. The side of the rectangular area that borders the river does not need fencing. The total area of the land is 1800 m^2. If it costs two dollars per meter of fencing, how can the farmer minimize the profits.
......y
....._____
....|......|
x..|.......| river
...|_____|
.......y

I know that Iv'e done this question a few time in math 12; Now i'm in my calculus class and understand the question. What im looking for is the minimum of the quadratic equation. However I can't seem to get the equation to be quadratic.

Here's what I got so far:

1800 = (x)(y) * my area equation

C = 2x + 2(2y) * my cost equation (2 dollars for every x amount of meters, and 2 dollars for every y amount of meters. The reason why I only have one 2 in front of my x in the equation is because there is only one "x side" but "2 y sides" - the other "x side" is technically bordered by the river.

I do some substitution and have this:

C = 2x = 2(2(1800/x))

This equation isn't quadratic so I know I missing something or done something wrong.

I appreciate any help.

 

[TchrWill]

Re: Extreme Values

A farmer wants to put a rectangular perimeter around some land that just so happens to by near a river. The side of the rectangular area that borders the river does not need fencing. The total area of the land is 1800 m^2. If it costs two dollars per meter of fencing, how can the farmer minimize the profits.
......y
....._____
....|......|
x..|.......| river
...|_____|
.......y
I know that Iv'e done this question a few time in math 12; Now i'm in my calculus class and understand the question. What im looking for is the minimum of the quadratic equation. However I can't seem to get the equation to be quadratic.
Here's what I got so far:
1800 = (x)(y) * my area equation
C = 2x + 2(2y) * my cost equation (2 dollars for every x amount of meters, and 2 dollars for every y amount of meters. The reason why I only have one 2 in front of my x in the equation is because there is only one "x side" but "2 y sides" - the other "x side" is technically bordered by the river.
I do some substitution and have this:
C = 2x = 2(2(1800/x))
This equation isn't quadratic so I know I missing something or done something wrong.
I appreciate any help


Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area

The traditional calculus approach would be as follows.

Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x^2.

Taking the first derivitive and setting equal to zero, dA/dx = P - 4x = 0, x becomes P/4.

With x = P/4, we end up with a rectangle with side ratio of 2:1.
.....The short side is P/4.
.....The long side is (P - 2(P/4)) = P/2.

Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, the long side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.

The corollary to this is that of all possible rectangles with a given area, the long side being a given external boundry, the rectangle with the side ratio of 2:1 has the smallest perimeter.

 

[Idealistic]

The thing is, is that im not looking for the highest area. The area is 1800 m^2, I need to find the smallest possible perimeter to encompass 1800 m^2 granted that one of the sides is bordered by a river. The perimeter costs the farmer 2 dollars per meter and I have to find out where the values for the sides are minimal.

 

[galactus]

The area is xy=1800

The perimeter is P=2x+y, but it costs $2 per foot, so we have 2(2x+y)

Solve the area equation for y and sub into the perimeter:

\(\displaystyle P=2(2x+\frac{1800}{x})\)

Now, this is what must be minimized. Differentiate, set to 0 and solve for x.

 

[TchrWill]

Sorry for the incorrect response.

 

[Idealistic]

The area is xy=1800

The perimeter is P=2x+y, but it costs $2 per foot, so we have 2(2x+y)

Solve the area equation for y and sub into the perimeter:

\(\displaystyle P=2(2x+\frac{1800}{x})\)

Now, this is what must be minimized. Differentiate, set to 0 and solve for x.

Thanx, but here's what happens if I assume y is the largest side like you have:

Cost = 4x + 3600/x (essentially starting where you left off)

C' = 4 - 3600/x^2

-4x^2 = -3600

x = + or - 30

Now If I where to assume that x was the largest side heres what I get:

Cost = 2x + 7200/x (this comes from P = x + 2y, then subbing from my area equaition, and then multiplying everying by two)

C' = 2 - 7200/x^2

-2x^2 = -7200

x = + or - 60

How do I know which is correct, the 30 or the 60?

 

[Deleted member 4993]

How do I know which is correct, the 30 or the 60?

Both are. They are describing similar rectangles - in one case x = 60 and y =30 ; the other x = 30 and y = 60