Hi Crystal,
Good question! I hope this explanation helps. :)
You know your farmer has 230 feet of fencing, so we're going to keep that.
You're looking for Area, so what is the area of a rectangle?
Area = x * y
But, first you need to define one of your variables in order to proceed. We do this by taking the perimeter of your rectangle. To get perimeter, you add the 4 sides, 2 of the length and 2 of the width.
Perimeter = 2x + 2y
Now, remember your 230 feet, that is the total perimeter you can possibly have because it's the maximum amount of fencing you have. Plug that in for P!
230 = 2x + 2y
Now, solve for one of your variables. Personally, I almost always solve for y because in a quadratic I prefer to work with x's.
So:
230 = 2x + 2y
-2x -2x
230 - 2x = 2y
_____________
2 (to get y alone)
115 - x = y
Great! Now you have defined one of your terms! You have a value for y. Plug that value for y in as y in your area formula, and solve.
A(x) = x * y
A(x) = x * (115 - x)
A(x) = 115x - x^2
You have a quadratic now:
A(x)= -x^2 + 115 x
Now, once you have it in this form, remember the form of a quadratic equation:
Ax^2 + Bx + C (A, B, and C are just your coefficients and they are integers)
To find your maximum area, you need to use this formula:
x = -B Here, B = 115
_____
2(A) Here, A = -1
So you have:
x = - 115
______
2 (-1)
x = -115
______
-2
A farmer has 230 ft of fence to enclose a rectangular garden. What is the largest garden area that can be enclosed with the 230 ft of fence? Explain your work.
10 answers
x = -115
____
-2
Solve this to get: 57.5
So, your greatest value for x will be 57.5.
Plug this in to your perimeter equation to determine the value of y.
Remember:
P(x) = 230 = 2x + 2y
230 = 2(57.5) + 2y
230 = 115 + 2y
-115 -115
___________________
115 = 2y
___ ___
2 2
57.5 = y
You have a sqaure! You now know that the value of x that will produce your maximum area is 57.5 and your value for y that will produce maximum area is 57.5.
Now, remember your area formula?
A(x) = x * y
Plug in your variables to find maximum area :)
A(x) = 57.5 * 57.5 = 3,306.25 Feet
Your maximum area = 3,306.25 Feet.
____
-2
Solve this to get: 57.5
So, your greatest value for x will be 57.5.
Plug this in to your perimeter equation to determine the value of y.
Remember:
P(x) = 230 = 2x + 2y
230 = 2(57.5) + 2y
230 = 115 + 2y
-115 -115
___________________
115 = 2y
___ ___
2 2
57.5 = y
You have a sqaure! You now know that the value of x that will produce your maximum area is 57.5 and your value for y that will produce maximum area is 57.5.
Now, remember your area formula?
A(x) = x * y
Plug in your variables to find maximum area :)
A(x) = 57.5 * 57.5 = 3,306.25 Feet
Your maximum area = 3,306.25 Feet.
37694046292365
P=2x+2y(perimeter)
JUST TO ADD TO THE ABOVE SOLUTION.
230=2x+2y
2y=230-2x
y=115-x
A=xy(Area)
A=x[115-x]
A=115x-x^2
dA/dX=115-2X=0(at maximum)
115=2x
x=57.5(CRITICAL VALUE)
JUST TO ADD TO THE ABOVE SOLUTION.
230=2x+2y
2y=230-2x
y=115-x
A=xy(Area)
A=x[115-x]
A=115x-x^2
dA/dX=115-2X=0(at maximum)
115=2x
x=57.5(CRITICAL VALUE)
105 feet by 105 feet
57.5
OBAMA IS A LIZARD SENT TO DESTROY EARTH
Yeaaaaaaa Boiiiiiiiiii
A farmer wants to know the area of a garden that is 312312 meters by 125125 meters. He uses the given model for help. Which expression can be used to find the area in square meters?
Hi Crystal,
Good question! I hope this explanation helps. :)
You know your farmer has 230 feet of fencing, so we're going to keep that.
You're looking for Area, so what is the area of a rectangle?
Area = x * y
But, first you need to define one of your variables in order to proceed. We do this by taking the perimeter of your rectangle. To get perimeter, you add the 4 sides, 2 of the length and 2 of the width.
Perimeter = 2x + 2y
Now, remember your 230 feet, that is the total perimeter you can possibly have because it's the maximum amount of fencing you have. Plug that in for P!
230 = 2x + 2y
Now, solve for one of your variables. Personally, I almost always solve for y because in a quadratic I prefer to work with x's.
So:
230 = 2x + 2y
-2x -2x
230 - 2x = 2y
_____________
2 (to get y alone)
115 - x = y
Great! Now you have defined one of your terms! You have a value for y. Plug that value for y in as y in your area formula, and solve.
A(x) = x * y
A(x) = x * (115 - x)
A(x) = 115x - x^2
You have a quadratic now:
A(x)= -x^2 + 115 x
Now, once you have it in this form, remember the form of a quadratic equation:
Ax^2 + Bx + C (A, B, and C are just your coefficients and they are integers)
To find your maximum area, you need to use this formula:
x = -B Here, B = 115
_____
2(A) Here, A = -1
So you have:
x = - 115
______
2 (-1)
x = -115
______
-2
Good question! I hope this explanation helps. :)
You know your farmer has 230 feet of fencing, so we're going to keep that.
You're looking for Area, so what is the area of a rectangle?
Area = x * y
But, first you need to define one of your variables in order to proceed. We do this by taking the perimeter of your rectangle. To get perimeter, you add the 4 sides, 2 of the length and 2 of the width.
Perimeter = 2x + 2y
Now, remember your 230 feet, that is the total perimeter you can possibly have because it's the maximum amount of fencing you have. Plug that in for P!
230 = 2x + 2y
Now, solve for one of your variables. Personally, I almost always solve for y because in a quadratic I prefer to work with x's.
So:
230 = 2x + 2y
-2x -2x
230 - 2x = 2y
_____________
2 (to get y alone)
115 - x = y
Great! Now you have defined one of your terms! You have a value for y. Plug that value for y in as y in your area formula, and solve.
A(x) = x * y
A(x) = x * (115 - x)
A(x) = 115x - x^2
You have a quadratic now:
A(x)= -x^2 + 115 x
Now, once you have it in this form, remember the form of a quadratic equation:
Ax^2 + Bx + C (A, B, and C are just your coefficients and they are integers)
To find your maximum area, you need to use this formula:
x = -B Here, B = 115
_____
2(A) Here, A = -1
So you have:
x = - 115
______
2 (-1)
x = -115
______
-2