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Section 4.2b: Maximization/Minimization
Applications of Quadratic Functions

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Objective o Interlude: maximization/minimization problems.

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Maximization/Minimization Problems
Many applications of mathematics involve determining the value (or values) of the variable x that returns either the maximum possible value or the minimum possible value of some function f(x). Such problems are called max/min
problems for short.

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Maximization/Minimization Problems
We are now able to solve max/min problems involving quadratic functions. If the parabola opens upward , we know that the vertex is the lowest (minimum) point on the graph. If the parabola opens downward , we know that the
vertex is the highest (maximum) point on the graph. Either way, we locate the vertex by completing the square.

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Maximization/Minimization Problems
We can shorten the process of locating the vertex by completing the square on the generic quadratic:
  
 ax
2
 a x

2 bx
 b a
 c x

 c
As always, we begin by factoring the leading coefficient a from the first two terms.

 a x
2  b b x a

4 a
2
2

 b
2
 
4 a c
To complete the square, we add the square of half of inside the a parentheses. This means we also




Vertex:

2
2 b a
2
 b 4 ac b
,
 a 4 a
4

2


4 a
2 have to subtract a times this quantity outside the parentheses.
Note: the vertex must lie at


 b
2 a
, f

 b
2 a 

 .

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Example 1: Maximization/Minimization Problems
Find the vertex of the following functions.
a.
  ver tex


3

 x

2  
1
,
7
    
 
1
2



1
,
85
6 12


Remember that the vertex lies at


 b 4 ac b
,

2 a 4 a
2


. b.
    vertex



5
 x
2
2

2

2
  x
,

4
4
    
4
 
2
2




1
,

19
5 5


Continued on the next slide…

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Example 1: Maximization/Minimization
Problems (cont.)
Find the vertex of the following function.
c.
  vertex


 x
2

2
3 x

3

1

,
6
4
    
4
 
2




3 33
,
2 4


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Example 2: Maximization/Minimization Problems
A farmer plans to use 2500 feet of spare fencing material to form a rectangular area for cows to graze against the side of a long barn, using the barn as one side of the rectangular area. How should he split up the fencing among the other three sides in order to maximize the rectangular area?
MOO!
x x
2500 2 x
If we let x represent the length of one side of the rectangular area then the dimensions of the rectangular area are x feet by 2500-2x feet (see image above). We will let A be the name of our function that we wish to maximize in this problem, so we want to find the
   
Continued on the next slide…

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Example 2: Maximization/Minimization Problems
(Cont.)
Note: If we multiply out the formula for A , we get a quadratic function.
   x

 
2 x
2 
2500 x
We know this function is a parabola facing down. We also know that the vertex is the maximum point on this graph.


Remember, the vertex is the point

2 b a
,
4 ac b
4 a
2

 o r



2 b a
, f


2 b a 


.
So, plugging in the values, we get the vertex


625, A
  
.
Therefore, the maximum possible area is A(625):
A (625)

781250 square fee t .