For what value(s) of x will the function f (x) = x ^{3} 2x ^{2}  5x be largest on the interval [ 10, 10] ?
This problem is asking us to find the absolute maximum on the interval. One way to do this is to graph the function and determine by inspection what the absolute maximum is. However, this is a very timeconsuming process. A more streamlined approach is to identify all possible candidates for a local maximum and compare them to each other to see which is the largest.
The possible places for an absolute maximum to occur are at the local maxima or at the endpoints of the interval. The first step is to identify the critical points and classify them as local maxima, local minima, or neither.
f'(x) = x ^{2}  4x  5 = (x  5)(x + 1), 
so
f'(x) = 0
at
x = 5
and
x =  1
.
Now, to classify these points, the second derivative test can be used.
f''(x) = 2x  4
;
f''(5) = 6
, which is greater than zero, so
f (5)
is a local minimum.
f''( 1) =  6
,which is less than zero, so
f ( 1)
is a local maximum.
Finally, the actual values of the local maxima and the endpoints should be compared to
each other to see which of these is the absolute maximum.
f ( 10) =  283
f ( 1) = 2
f (10) = 83
So, the absolute maximum on the interval occurs when
x = 10
.
Example: For what value(s) of
x
will the function
f (x) =
x
^{3} 2x
^{2}  5x
be largest for
all
x
?
This problem is identical to the one above, except that there are no endpoints to compare to the local maxima.
In this situation, we must check what happens to the function as x approaches positive and negative infinity. By inspection, it becomes clear that as x approaches positive infinity, f also approaches positive infinity. Thus, the function grows without bound, and there is no absolute maximum.
A builder needs to make a box with a square bottom and rectangular sides. The box has no top. If the material for the sides cost $2 per square foot, and the material for the bottom costs $4 per square foot, what is the largest volume box that the builder can make with $20?
This problem is known as a "constrained optimization" problem. The procedure for solving this sort of problem is ultimately similar to the procedure described above for optimizing functions of one variable. However, some work is required to transform this word problem into a function of one variable. The first three steps below describe this process.
Step One: Identify the objective function and express it in terms of the
relevant variables.
The objective function represents the quantity that is ultimately going to be maximized or
minimized. In this case, the quantity of interest is the volume of the box, and it needs to
be maximized. The relevant variables here are the dimensions of the box. It is often
useful to draw a diagram:
Let
x
be the both the length and width of the square bottom of the box.
Let
y
be the height of the sides of the box.
Expressing the volume in terms of the relevant variables generates the objective function: V = x ^{2} y . This quantity must be maximized.
Step Two: Identify the constraint.
The constraint is the rule or equation that relates the variables used to generate the objective function. In this case, the way to relate the variables x and y is to use the fact that the total price of the box materials must equal $20. Since the cost of the material is the area of the material multiplied by the cost per square foot, the constraint can be expressed as follows:
(4xy)(2) + (x ^{2})(4) = 20
Step Three: Use the constraint to express the objective as a function of one
variable.
The methods that we have learned to analyze functions only apply to functions of one variable. The constraint can be used to reduce the objective to a function of one variable so that our techniques of finding maxima and minima will apply. This involves using the constraint to solve for one variable in terms of another. In this case, we solve for y , although solving for x will work also:
y = =  <BR>
Now, this can be substituted back into the original objective to yield:
V = x ^{2}   

Step Four: Now,
V
is expressed as a function of one variable,
x
, and
procedures explained previously to for optimizing functions of one variable can be used.
The domain of V(x) is (0, + ∞) . This is because x could never be a negative quantity, and could not be zero.
V'(x)  =  x ^{2}  
V'(x)  = 0 whenx = ± 
but only x = + is in the domain of V.
Now, to check if this critical point is a local maximum, minimum, or neither, the second derivative test can be used:
V''(x) =  3x  
V'' =  3 < 0 
Because the second derivative is negative, this critical point is a local maximum.
We can also be sure that this is the absolute maximum on the open interval (0, + ∞) . This is because there are there are no more critical points on this interval, so the graph must only be increasing to the left of the critical point, and decreasing to the right. To answer the original problem, the largest possible volume is:
V  =   
=   =  
= square feet 