```HAWKES LEARNING SYSTEMS
math courseware specialists
Hawkes Learning Systems:
College Algebra
Section 4.2b: Maximization/Minimization
HAWKES LEARNING SYSTEMS
math courseware specialists
Objective
o Interlude: maximization/minimization problems.
HAWKES LEARNING SYSTEMS
math courseware specialists
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.
HAWKES LEARNING SYSTEMS
math courseware specialists
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.
HAWKES LEARNING SYSTEMS
math courseware specialists
Maximization/Minimization Problems
We can shorten the process of locating the vertex by
completing the square on the generic quadratic:
f  x   ax 2  bx  c
As always, we begin by factoring
 2 b 
the leading coefficient a from the
 a x  x  c
a 
first two terms.

To complete the square, we add
 2 b
b2  b2
b
the
square
of
half
of
inside the
 a x  x  2  
c
a
a
4
a
4
a
parentheses. This means we also


b  4ac  b 2

 a x   
2a 
4a

 b 4ac  b 2 
Vertex:   ,

2
a
4
a


2
have to subtract a times this
quantity outside the parentheses.
Note: the vertex must lie at
 b
 b 

,
f
 2a   2a  .



HAWKES LEARNING SYSTEMS
math courseware specialists
Example 1: Maximization/Minimization Problems
Find the vertex of the following functions.
2
f
x

3
x
 x7
a.  
 1 4  3 7   1 2   1 85 
   , 
vertex   
,

4  3
 23
  6 12 
Remember that
the vertex lies at
 b 4ac  b 2 
 ,
.
2
a
4
a


b. f  x   5 x 2  2 x  4
2

2 4  5 4   2   1 19 
  , 
vertex   
,

4  5
 2  5
 5 5 
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 1: Maximization/Minimization
Problems (cont.)
Find the vertex of the following function.
2
f
x


x
 3x  6
c.  
2

3 4  1 6   3   3 33 
  , 
vertex   
,

4  1
 2  1
  2 4
HAWKES LEARNING SYSTEMS
math courseware specialists
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  2x
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
maximum possible value of A  x   x  2500  2 x  .
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 2: Maximization/Minimization Problems
(Cont.)
Note: If we multiply out the formula for A, we get a
quadratic function. A  x   x  2500  2 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
 b 4ac  b2 
 b
 b 
  2a , 4a  or   2a , f   2a   .





So, plugging in the values, we get the vertex   625, A  625 .
Therefore, the maximum possible area is A(625):
A(625)  781250 square feet.
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