Use the formula for the axis of symmetry to get the x-coordinate for the vertex.
Width with Maximum Area: 25 ft Maximum Area: 625 ft^2 feet Range: 0 < A ≤ 625
Practice makes perfect
This exercise asks us to complete three tasks with the function A=- x^2+50x.
What is the width of its maximum area?
What is its maximum area?
What is the range?
In order to find maximums and minimums of a standard quadratic functions, y= ax^2+ bx+c, we will need to use the equation for the axis of symmetry.
x=- b/2 a
Then once we have the axis of symmetry, we can substitute the x-value into our original equation to find the maximum. The maximum will give us the upper bound for the range of our function.
Width
Let's start by comparing our function to the standard quadratic.
ccc
y&=& ax^2&+& bx&+&c
↕ & & ↕ & & ↕ & & ↕
A&=& -x^2&+& 50x&+&0
We can see that x is our independent variable that defines our width. The area, A, is our dependent variable and represents y in the standard form. Then we can see that a = -1, b= 50, and c= .
Now that we know the maximum, we know the area cannot be any greater than 625ft^2. In general, measurements are not negative, so the minimum area of the rectangle is 0, and we can say that the range is 0 < A ≤ 625.
