a rancher wants to use 300 feet of fencing to enclose a rectangular area of 4400 square feet. what dimensions should the rectangle have?
length = x
width = y
x+y = 150 , y = 150-x
xy = 4400
x(150-x) = 4400
expanding and simplifying...
x^2 - 150x + 4400 = 0
(x-110)(x-40) = 0
x = 110 or x=40
if x=110, then y = 40
if x = 40, then y = 110
the dimenstions should be 110 ft by 40 ft
check: 2(110+40) = 300
(110)(40) = 4400
To find the dimensions of the rectangle, we need to use the given information.
Let's assume the length of the rectangular area is 'l' feet and the width is 'w' feet.
The formula to calculate the perimeter (the amount of fencing needed) of a rectangle is given by:
Perimeter = 2 * (Length + Width)
In this case, the perimeter is given as 300 feet. So we can write the equation as:
300 = 2 * (l + w)
Simplifying the equation, we get:
l + w = 150
Next, we have been provided with the area of the rectangle, which is 4400 square feet. The formula to calculate the area of a rectangle is given by:
Area = Length * Width
In this case, the area is given as 4400 square feet. So we can write another equation as:
l * w = 4400
We now have a system of two equations:
l + w = 150 (Equation 1)
l * w = 4400 (Equation 2)
To solve this system of equations, we can use substitution or elimination method. Let's solve it using the substitution method:
From Equation 1, we can rewrite it as:
l = 150 - w
Substituting this value of 'l' in Equation 2, we get:
(150 - w) * w = 4400
Expanding and rearranging the equation, we get:
w^2 - 150w + 4400 = 0
Now, we can solve this quadratic equation to find the possible values of 'w'. We can factorize the equation or use the quadratic formula. Let's solve it using the quadratic formula:
The quadratic formula is given by:
w = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -150, and c = 4400.
Substituting these values in the quadratic formula, we get:
w = (-(-150) ± √((-150)^2 - 4 * 1 * 4400)) / (2 * 1)
Simplifying further, we get:
w = (150 ± √(22500 - 17600)) / 2
w = (150 ± √4900) / 2
w = (150 ± 70) / 2
Now, we have two possible values for 'w':
w1 = (150 + 70) / 2 = 220 / 2 = 110
w2 = (150 - 70) / 2 = 80 / 2 = 40
Thus, we have two possible widths: 110 feet and 40 feet.
To find the corresponding lengths, we substitute these values of 'w' back into Equation 1:
For w = 110:
l + 110 = 150
l = 150 - 110
l = 40
So, one possible set of dimensions is length = 40 feet and width = 110 feet.
For w = 40:
l + 40 = 150
l = 150 - 40
l = 110
The other possible set of dimensions is length = 110 feet and width = 40 feet.
Therefore, the rectangle can have dimensions of either 40 feet by 110 feet or 110 feet by 40 feet to enclose a rectangular area of 4400 square feet with 300 feet of fencing.