2 more problems -
Factory is to be bulit on a lot 180 x 240 . Local building code - specifies that lawn uniform width and equal in area of the factory which it surrounds. What must the width of this lawn be? Dimensions of the factory.
Let x be the required width. The factory area will be
(240-2x)(180-2x). That must equal half of (240x180), which is 21,600.
Solve the quadratic equation
(240-2x)(180-2x)= 21,600
It can be rewritten as
(120-x)(90-x) = 5400
or
x^2 - 210 x + 5400 = 0
(x-30)(x-180) = 0
There are two roots, but only one makes sense. A lawn 180 m wide would overlap and leave no room for the factory.
To answer this question, we need to break it down into smaller parts. Let's first determine the dimensions of the factory.
Given that the lot is 180 x 240, it means that the total area of the lot is 180 * 240 = 43,200 square units.
Let's assume the width of the lawn is 'x.' Since the lawn surrounds the factory uniformly, it means that there will be lawn on all four sides of the factory.
So, the dimensions of the factory would be:
Length = 180 - 2x
Width = 240 - 2x
Next, we know that the lawn should have the same area as the factory. Since the lawn is rectangular, its area is given by:
Lawn Area = x * (180 - 2x) * x * (240 - 2x)
According to the given condition, the lawn's area should be equal to the factory's area, which is 43,200 square units.
Therefore, we can set up the following equation:
x * (180 - 2x) * x * (240 - 2x) = 43,200
Now, we can solve this equation to find the value of 'x,' which represents the width of the lawn. However, solving this equation is a bit complex and involves polynomial equations.
To find the value for 'x,' we can use numerical methods or graphing calculators to solve the equation. One way is to use a graphing calculator or software that can plot the equation y = x * (180 - 2x) * x * (240 - 2x) and find the x-coordinate(s) where y = 43,200.
Alternatively, we can approximate the solution using numerical methods like the Newton-Raphson method or trial and error method. In these methods, we make a guess for the value of 'x' and refine it until we get close to the solution.
Once we find the value for 'x,' we can substitute it back into the expressions for the length and width of the factory to get the dimensions.