How to find dimensions of a field when given length of fencing and area?
Hi I really appreciate your help on this.
The question states that I have a field of area 650 square metres and fencing of 110 metres. I must give 3 possible sets of dimensions of the field and the fencing need not cover all four sides of the field.
Oh I forgot to mention that quadratic algebra and graphing utility must be used.
Thank you so much.
A = Area
F = Fencing
W = Width
L = Length
A = W * L = 650 m ^ 2
W * L = 650 Divide both sides by L
W = 650 / L
F = 2 W + 2 L = 2 ( W + L ) = 110 m
2 ( W + L ) = 110 Divide both sides by 2
W + L = 110 / 2
W + L = 55
650 / L + L = 55 Multiply both sides by L
650 + L * L = 55 L
650 + L ^ 2 = 55 L
L ^ 2 - 55 L + 650 = 0
The exact solutions are :
L = ( 5 / 2 ) * [ 11 + sqrt ( 17 ) ]
approx. 37.81 m
and
L = ( - 5 / 2 ) * [ - 11 + sqrt ( 17 ) ]
approx. 17.19 m
When L = 37.81 m
W = 650 / 37.81 = 17.19 m
When L = 17.19 m
W = 650 / 17.19 = 37.81 m
Dimensions of a field:
37.81 m X 17.19 m
If you don't know how solve equation L ^ 2 - 55 L + 650 = 0
in google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve by Quadratic Formula
When page be open in rectangle type:
L ^ 2 - 55 L + 650 = 0
and click option: solve it!
You wil see solution step-by-step
If you must draw function :
L ^ 2 - 55 L + 650 = 0
In google type:
functions graphs online
When you see list of results click on:
rechneronline.de/function-graphs/
When page be open in blue rectangle type:
x ^ 2 - 55 x + 650 = 0
Set:
Range x-axis from - 20 to 80
Range y-axis from - 200 to 800
And click option: Draw
CORRECTION :
When page be open in blue rectangle type:
x ^ 2 - 55 x + 650
Set:
Range x-axis from - 20 to 80
Range y-axis from - 200 to 800
And click option: Draw
To find the dimensions of a field when given the length of fencing and the area, you can set up a quadratic equation.
Let's denote the length of one side of the field as 'x' and the length of the adjacent side as 'y'.
1. Start by setting up the equation for the perimeter using the given length of fencing. Since the fencing doesn't have to cover all four sides, we can have various combinations of sides. One possible equation is:
Perimeter = 2x + y = 110
2. Next, set up the equation for the area using the given area of the field, which is 650 square meters:
Area = x * y = 650
3. Now we have a system of equations:
2x + y = 110
x * y = 650
To solve this system of equations using quadratic algebra and graphing utility, follow these steps:
a. Rewrite the first equation in terms of one variable (y) using the second equation:
y = 110 - 2x
b. Substitute this value of y into the second equation:
x(110 - 2x) = 650
c. Simplify the equation:
-2x^2 + 110x - 650 = 0
d. Use the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -2, b = 110, and c = -650.
e. Calculate the two possible values of x and substitute them back into the equation y = 110 - 2x to find the corresponding values of y.
f. Repeat the steps above to find two more possible sets of dimensions for the field, keeping in mind that x and y can be swapped to find different combinations.
Remember to use a graphing utility to visualize the quadratic equation, which can help verify the solutions and find any additional possible dimensions of the field.
Overall, these steps should help you find three possible sets of dimensions for the field given the length of fencing and area, using quadratic algebra and a graphing utility.