Try our Free Online Math Solver!

Optimization
Example 5. Use the formula
to find the x coordinate of the
vertex of the parabola represented by the quadratic function in Example 2.
In Example 2, the quadratic function was represented by the equation
f(x) = 2x^{2} + 12x + 12.
In vertex form
f(x) = 2(x + 3)^{2} − 6,
the coordinates of the vertex were easily seen to be
(−3,−6) (see Figure 2). Let’s see
what the new formula for the x coordinate of the vertex reveals.
As usual, compare f (x) = 2x^{2} + 12x + 12 with f(x) = ax^{2}
+ bx + c and note that
a = 2, b = 12 and c = 12. Thus, the xcoordinate of the vertex is given by
Note that this agrees with the previous result (see Figure
2). We could find the
ycoordinate of the vertex with
but we find this formula for the ycoordinate of the
vertex a bit hard to memorize.
We find it easier to do the following. Since we know the xcoordinate of the
vertex is
x = −3, we can find the ycoordinate of the vertex by simply substituting x = −3
in
the equation of the parabola . That is, with f(x) = 2x^{2} + 12x − 12,
f(−3) = 2(−3)^{2} + 12(−3) + 12 = −6.
Let’s highlight this last technique.
Finding the ycoordinate of the Vertex. Given the parabola
represented by
the quadratic function
f(x) = ax^{2} + bx + c,
we’ve seen that the xcoordinate of the vertex is given by
x = −b/(2a). To find
the ycoordinate of the vertex, it is probably easiest to evaluate the function
at
x = −b/(2a). That is, the ycoordinate of the vertex is given by
Let’s look at another example.
Example 6. Consider the parabola having equation
f(x) = −2x^{2} + 3x − 8.
Find the coordinates of the vertex.
First, use the new formula to find the xcoordinate of the vertex.
Next, substitute x = 3/4 to find the corresponding ycoordinate.
Thus, the coordinates of the vertex are (3/4,−55/8).
Applications
We’re now in a position to do some applications of
optimization. Let’s start with an
easy example.
Example 7. Find two real numbers x and y that sum to 50
and that have a
product that is a maximum .
Before we apply the theory of the previous examples, let’s
just play with the numbers
a bit to get a feel for what we are being asked to do. We need to find two
numbers that
sum to 50, so let’s start with x = 5 and y = 45. Clearly, the sum of these two
numbers
is 50. On the other hand, their product is xy = (5)(45) = 225. Let’s place this
result
in a table.
For a second guess, select x = 10 and y = 40. The sum of
these two numbers is 50
and their product is xy = 400. For a third guess, select x = 20 and y = 30. The
sum
of these two numbers is 50 and their product is xy = 600. Let’s add these
results to
our table.
Thus far, the best pair is x = 20 and y = 30, because
their product is the maximum
in the table above. But is there another pair with a larger product? Remember
our
goal is to find a pair with a product that is a maximum. That is, our pair must
have a
product larger than any other pair. Can you find a pair that has a product
larger than
600?
Now that we have a feel for what we are being asked to do
(find two numbers that
sum to 50 and that have a product that is a maximum), let’s try an approach that
is
more abstract than the “guess and check” approach of our tables. Our first
constraint is
the fact that the sum of the numbers x and y must be 50. We can model this
constraint
with the equation
x + y = 50. (8)
We’re being asked to maximize the product. Thus, you want
to find a formula for the
product. Let’s let P represent the product of x and y and write
P = xy. (9)
Note that P is a function of two variables x and y .
However, all of our functions in
this course have thus far been a function of a single variable. So, how can we
get rid
of one of the variables? Simple, first solve equation (8) for y.
x + y = 50
y = 50 − x (10)
Now, substitute equation (10) into the product in equation (9).
P = x(50 − x),
or, equivalently ,
P = −x^{2} + 50x. (11)
Note that P is now a function of a single variable x. Note
further that the function
defined by equation (11) is quadratic. If we compare P = −x^{2}+50x with the
general
form P = ax^{2} + bx + c, note that a = −1 and b = 50 (we have no need of the fact
that
c = 0). Therefore, if we plot P versus x, the graph is a parabola that opens
downward
(see Figure 3) and the maximum value of P will occur at the vertex. The
xcoordinate
of the vertex is found with
Figure 3. The maximum product
P=625 occurs at the vertex of the
parabola, (25, 625).
Thus, our first number is x = 25. We can find the second
number y by substituting
x = 25 in equation (10).
y = 50 − x = 50 − 25 = 25.
Note that the sum of x and y is x+y = 25+25 = 50. There
are two ways that we can
find their product. Since we now know the numbers x and y, we can multiply to
find
P = xy = (25)(25) = 625. Alternatively, we could substitute x = 25 in equation
(11)
to get
P = −x^{2} + 50x = −(25)^{2} + 50(25) = −625 + 1250 = 625.
When you compare this result with our experimental tables,
things come together.
We’ve found two numbers x and y that sum to 50 with a product that is a maximum.
No other numbers that sum to 50 have a larger product.
Our little formula has
proven to be a powerful ally . Let’s try
another example.
Prev  Next 