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Hawkes Learning Systems:
College Algebra
Section 4.2b: Maximization/Minimization
Applications of Quadratic Functions
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Objective
o Maximization/minimization problems.
HAWKES LEARNING SYSTEMS
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Copyright © 2011 Hawkes Learning Systems.
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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 or
using the formula.
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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  .



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Example: 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…
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Example: 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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Example: Maximization/Minimization
Problems
a) The total revenue of Machinery Rental is given by
the function
where x is
the number of units rented. What number of units
rented produces the maximum revenue?
b) The total cost of manufacturing a set of golf clubs is
given by
where x is
the number of sets of golf clubs produced. How
many sets of golf clubs should be made to incur
minimum cost?
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2011 Hawkes Learning Systems.
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Example: Maximization/Minimization
Problems
Height in feet at time t may be found with this function:
Sitting in a tree, 48 feet above ground level, Sue shoots
a pebble straight up with a velocity of 64 feet per
second. What is the maximum height attained by the
pebble?
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Slides for 4.2b