Characteristics of Quadratic Functions Example
I do not understand step#4. Where did the -w come from? Everything else I understand competely.
A rectangular field is to be enclosed by 600m of fence.
- What is the maximum area?
- What dimensions will give a maximum area?
Step by Step...
Let l be the length of the rectangle
Let w be the width of the rectangle
Let A be the area of the rectangle.
Draw a diagram if possible, and label with appropriate variables
Write down the equations.
The 600m fencing means the perimeter of the rectangle is 600 m. Therefore, the first equation is:
2l + 2w = 600 ---- (1)You were asked to maximize the area of the rectangle, so the second equation is:
A = l × w --- (2)
Get a quadratic function.
Use equation (1) to solve for or w. Let’s say you solve for l.
l = −w + 300 ---- (3)
You need to remember the equation that gives the area of rectangle, such that:
`A = l*w`
l represents the length
w represents the width
Since the problem provides the value of perimeter of rectangle, you may write area, either in terms of length, or in terms of width, using the equation of perimeter, such that:
`600 = 2(l + w) => 300 = l + w => l = 300 - w`
Hence, substituting back 300 - w for l in equation of area yields:
`A(w) = (300 - w)*w`
Opening the brackets yields:
`A(w) = 300w - w^2`
You need to maximize the area of rectangle, hence, you need to differentiate the function `A(w) = 300w - w^2` with respect to w, such that:
`A'(w) = (300w - w^2)' => A'(w) = 300*w' - (w^2)'`
`A'(w) = 300*1 - 2w => A'(w) = 300 - 2w`
Now, you need to solve for w the equation `A'(w) = 0` such that:
`300 - 2w = 0 => 2w = 300 => w = 150 m`
Hence, evaluating the dimensions that maximize the area of the fence, yields `w=l = 150 m` .
Now, substitute – w + 300 for l into equation (2)
A = (−w + 300) × w
= (−w2 + 300w) this is your quadratic function.