Find 2 positive numbers whose product is 25 and whose sum is a minimum.
Objective: S = x + y
. The goal is to minimize S
Constraint: xy = 25
Substitute constraint into objective:
|S|| ||= + x; domain = (0,∞)||
|S'(x)|| ||= + 1||
| || ||S'(x) = 0 when x = 5||
Use second derivative to classify:
S''(5) > 0
|S''(x) = ||
, so S
has a local min at x = 5
However, notice that S''(x)
is always positive on the
, so S
is always concave up on that interval,
which means that the local min is also the absolute min.
Therefore, 5 and 5 are the positive numbers with the smallest
sum whose product is 25.
What is the maximum value of f (x) = x4 -8x2 - 3 on the interval [- 3, 3]?f'(x) = 4x3 -16x = 4x(x2 - 4)
This equals zero when x
equals 0 or -2 or +2.
Making use of the sign of the first derivative yields the following chart for the behavior
Based on this, the only local maximum occurs at x = 0
. Now, to see if this is an
absolute maximum on the interval, compare the value of f
at x = 0
to the value at the
f (- 3) = 6f (0) = - 3f (3) = 6
While x = 0
is a local maximum, it is not the absolute maximum on this interval. The
absolute maximum occurs at both of the endpoints in this case.
A shepherd wishes to build a rectangular fenced area against the side of a barn. He has 360 feet
of fencing material, and only needs to use it on three sides of the enclosure, since the wall
of the barn will provide the last side. What dimensions should the shepherd choose to
maximize the area of the enclosure?
Below is a sketch of the situation:
Objective: maximize A = xy
Constraint: 2y + x = 360
Substitution into objective: A = (360 - 2y)(y)A(y) = 360y - 2y2A'(y) = 360 - 4yA'(y) = 0
at x = 90A''(y) = - 4
, so A''(90) < 0
and y = 90
is a local maximum.
However, because A''(y) = - 4
for all y
, the graph of A(y)
is always concave down,
so the local maximum is also the absolute maximum. Thus, choosing y = 90
x = 180
ft will generate the largest area.
Find the point on the graph of y = x2 that is the smallest distance from the point
Objective: Let D
= the distance between the point (x,y) on the graph of f (x) = x2
the point (0, 6)
We want to minimize this function.
In practice, minimizing the distance is the same as minimizing the square of the distance,
and because working with square roots can become complicated, we will choose here to
minimize the function D2
, which is the square of the distance.
So, D2 = (y - 6)2 + x2
Constraint: y = x2
Substituted objective: D2 = (x2 -6)2 + x2
| || ||(D2)'(x) = 2(x2 -6)2(2x) + 2x||
| || ||= 4x3 - 24x + 2x||
| || ||= x(4x2 - 22)||
| || ||(D2)'(x) = 0 atx = 0 and atx = ±||
| || ||Now use the second derivative:||
| || ||(D2)''(x) = 12x2 - 22||
| || ||(D2)''(0) < 0, so it is a local max.||
| || ||(D2)''() > 0, so it is a local min.||
| || ||(D2)''(- ) > 0, so it is a local min also.||
Note that by symmetry, the points (,)
on the graph of y = x2
the exact same distance from the point (0, 6)
. To see that these local minima
are also at the absolute minimum distance, consider the following diagram.
From this information, it should be apparent that this function
attains its lowest values at x = ±.