§4.2 Optimization.
The student will learn about
the optimization of
a function in
several different
settings.
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Reminder of Facts
f is increasing
f ‘ (x) > 0
f is decreasing
f ‘ (x) < 0
f is constant
f ‘ (x) = 0
f is concave up
f “ (x) > 0
f concave down
f “ (x) < 0
Inflection point
f “ (x) = 0
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Example 1
Find the max and min values of
f (x) = x3 – 12 x
f ’ (x) = 3x 2 - 12
so that the critical values are +2, -2.
Test at – 2 and 2, the critical values.
f (- 2) = 16
a max.
f ( 2) = - 16
a min.
I’m using my calculat
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Example 1 Continued
Find the intervals where the graph of f is concave
upward, the intervals where the graph of f is
concave downward if
f (x) = x3 – 12 x.
f ‘ (x) = 3x2 - 12
f “ (x) = 6x
f “ (x) is positive when x > 0
so f is concave up on the
region (0, ∞)
f ” (x) is negative when x < 0 so f it is concave
down on the region (∞, 0).
Check with your calculator.
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Example 1 Continued
Find the inflection points of
f (x) = x3 – 12 x.
f ‘ (x) = 3x2 - 12
f “ (x) = 6x
f “ (x) is zero when x = 0 so
there is an inflection point at
(0, 0).
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A new problem is coming up.
Clear your heads of the old one.
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Another Optimization Example
A peach grower finds that if he plants 40 trees per acre, each tree
will yield 60 bushels of peaches. He also estimates that for each
additional tree that he plants per acre, the yield will decrease by
2 bushels. How many trees per acre should he plant?
What should we let x equal?
Let x be the number of additional trees per acre.
The number of trees per acre t (x) = 40 + x.
The yield per tree y (x) = 60 - 2x.
The total yield Y (x) = (40 + x) (60 - 2x) = 2400 – 20x – 2x 2
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Another Example - Continued
P. 216, #6. A peach grower finds that if he plants 40 trees per
acre, each tree will yield 60 bushels of peaches. He also estimates
that for each additional tree that he plants per acre, the yield
will decrease by 2 bushels. How many trees per acre should he
plant?
The total yield Y (x) = (40 + x) (60 - 2x) = 2400 – 20x – 2x 2
Maximize Y with your
calculator
x = - 5 [35 trees/acre] and a
yield of y = 2450 bushels.
- 30 ≤ x ≤ 30 0 ≤ y ≤ 2600
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A new problem is coming up.
Clear your heads of the old one.
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Maximizing Volume
Suppose you have a 9“ by 12” piece of construction paper.
If you cut a square out of each corner and fold the sides up
you will have a box that is open at the top. What size square
should be cut out in order to maximize the volume of the
box?
Let x be the side of the square cut
out of the corner.
What is the volume formula?
x
V = lwh
Notice that this is four variable and we may only have
two and x and a V. What should we do?
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Continued
Maximizing Volume
Suppose you have a 9“ by 12” piece of construction paper. If you cut a
square out of each corner and fold the sides up you will have a box
that is open at the top. What size square should be cut out in order to
maximize the volume of the box?
V = lwh
l = 12 – 2x and w = 9 – 2x and h = l.
x
V = (12 – 2x)(9 – 2x)(x) = 108x - 42x 2 + 4x 3 .
We will now set the first derivative equal to zero and
solve.
V ‘ = 108 – 84x + 12x 2 = 12 ( 9 – 7x + x 2)
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Continued
Maximizing Volume
Suppose you have a 9“ by 12” piece of construction paper. If you cut a
square out of each corner and fold the sides up you will have a box
that is open at the top. What size square should be cut out in order to
maximize the volume of the box?
V ‘ = 108 – 84x + 12x 2 = 12 ( 9 – 7x + x 2)
x
Using the quad formula or your
calculator yields x = 5.30 and x = 1.70.
Notice that the first solution is not possible. Why?
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Summary.
• We did an optimization problem using a cubic
function.
• We optimized a peach harvest.
• We optimized a volume.
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ASSIGNMENT
§4.2; Page 76; 1 – 19.
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