Respuesta :
Answer:
Step-by-step explanation:
Assuming the parking lot is along both widths of the field and only once along the length:
the length is (300 - 2x)
the width is (200 -x)
the area is (300-2x)(200-x) = 30000
60000 -300x - 400x +2x^2 = 30000
2x^2 -700x + 30000 = 0
x^2 - 350x + 15000 = 0
(x - 300 )(x - 50 ) = 0
x = 300 or x=50
x=300 results in negative measures.
the parking lot is x=50 wide
the length of the field is 200, the width of the field is 150
By solving a system of equations, we will see that are two possible solutions for the width:
x = 75m
x = 100m
How wide is the parking lot?
Remember that for a rectangle of length L and width W, the area is:
A = L*W
Here we know the length of the field and the parking lot, that we can write as:
L' = L + 2x = 300m
x is the width of the parking lot.
The width is:
W' = W + x = 200m
(here x appears only once because we know that the parking lot only surrounds 3 sides of the field).
And the area of the field is 30,000m^2, then:
W*L = 30,000m^2
So we have 3 equations:
L + 2x = 300m
W + x = 200m
W*L = 30,000m^2
This is a system of equations.
Using the third one, we can isolate W to get:
W = (30,000m^2)/L
Now we can replace this on the second equation to get:
(30,000m^2)/L + x = 200m
Now we need to isolate L (because we want a solution for x), so we get:
30,000m^2 = (200m - x)*L
(30,000m^2)/(200m - x) = L
Now we can replace this on the last equation to get:
L + 2x = 300m
(30,000m^2)/(200m - x) + 2x = 300m
Now we can solve this for x.
30,000m^2 + 2x*(200m - x) = 300m*(200m - x)
- 4x^2 + 700m*x -30,000m^2 = 0
So we have a quadratic equation, the solutions are given by Bhaskara's formula:
[tex]x = \frac{-700m \pm \sqrt{(700m)^2 - 4*(-4)*-30,000m^2} }{2*-4} \\\\x = \frac{-700m \pm 100m }{-8}[/tex]
So there are two possible solutions:
x = (-700m + 100m)/-8 = 75m
x = (-700m - 100m)/-8 = 100m
Which will depend on the values we take for L and W.
If you want to learn more about systems of equations, you can read:
https://brainly.com/question/13729904
