Download Section 4.7

APPLICATIONS OF DIFFERENTIATION
4.7
Optimization Problems
In this section, we will learn:
How to solve problems involving
maximization and minimization of factors.
OPTIMIZATION PROBLEMS
In this section (and the next), we solve
such problems as:
 Maximizing areas, volumes, and profits
 Minimizing distances, times, and costs
OPTIMIZATION PROBLEMS
Example 1
A farmer has 2400 ft of fencing and wants
to fence off a rectangular field that borders
a straight river. He needs no fence along
the river.
 What are the dimensions of the field that
has the largest area?
OPTIMIZATION PROBLEMS
Here are three
possible ways of
laying out the 2400 ft
of fencing.
Example 1
OPTIMIZATION PROBLEMS
Example 1
This figure
illustrates
the general case.
We wish to maximize the area A of
the rectangle.
 Let x and y be the depth and width of the rectangle
(in feet).
 Then, we express A in terms of x and y: A = xy
OPTIMIZATION PROBLEMS
Example 1
We want to express A as a function of
just one variable.
 So, we eliminate y by expressing it in terms of x.
 To do this, we use the given information that
the total length of the fencing is 2400 ft.
 Thus,
2x + y = 2400
OPTIMIZATION PROBLEMS
Example 1
From that equation, we have:
y = 2400 – 2x
This gives:
A = x(2400 – 2x) = 2400x - 2x2
 Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0).
OPTIMIZATION PROBLEMS
Example 1
So, the function that we wish to maximize
is: A(x) = 2400x – 2x2
0 ≤ x ≤ 1200
 The derivative is: A’(x) = 2400 – 4x
 So, to find the critical numbers, we solve: 2400 – 4x = 0
 This gives: x = 600
OPTIMIZATION PROBLEMS
Example 1
The maximum value of A must
occur either at that critical number or
at an endpoint of the interval.
 A(0) = 0; A(600) = 720,000; and A(1200) = 0
 So, the Closed Interval Method gives the maximum
value as:
A(600) = 720,000
OPTIMIZATION PROBLEMS
Example 1
Thus, the rectangular field should
be:
 600 ft deep
 1200 ft wide
OPTIMIZATION PROBLEMS
Example 2
A cylindrical can is to be made to
hold 1 L of oil.
 Find the dimensions that will minimize
the cost of the metal to manufacture the can.
OPTIMIZATION PROBLEMS
Draw the diagram as in
this figure, where
r is the radius and h the
height (both in
centimeters).
Example 2
OPTIMIZATION PROBLEMS
To minimize the cost of
the metal, we minimize
the total surface area of
the cylinder (top, bottom,
and sides.)
Example 2
OPTIMIZATION PROBLEMS
So, the surface area is:
A = 2πr2 + 2πrh
Example 2
OPTIMIZATION PROBLEMS
Example 2
To eliminate h, we use the fact that
the volume is given as 1 L, which we take
to be 1000 cm3.
 Thus,
πr2h = 1000
 This gives
h = 1000/(πr2)
OPTIMIZATION PROBLEMS
Example 2
Substituting this in the expression for A gives:
2000
 1000 
2
A  2 r  2 r 
 2 r 
2 
r
 r 
2
So, the function that we want to minimize is:
2000
A(r )  2 r 
r
2
r 0
OPTIMIZATION PROBLEMS
Example 2
To find the critical numbers, we differentiate:
2000 4( r  500)
A '(r )  4 r  2 
2
r
r
3
3
 Then, A’(r) = 0 when πr = 500
 So, the only critical number is: r  3 500 / 
OPTIMIZATION PROBLEMS
Example 2
2000 4( r  500)
A '(r )  4 r  2 
2
r
r
3
 We can observe that A’(r) < 0 for r  3 500 / 
and A’(r) > 0 for r  3 500 / 
 So, A is decreasing for all r to the left of the critical
number and increasing for all r to the right.
 Thus, r 
minimum.
3
500 /  must give rise to an absolute
OPTIMIZATION PROBLEMS
Example 2
The value of h corresponding to
r  500 /  is:
3
1000
1000
h

2
r
 (500 /  ) 2 3
500 

3


1000
  


2
500 

500


3
3



  
  
500 

1000 3 

500
  

 23
 2r
500



OPTIMIZATION PROBLEMS
Example 2
Thus, to minimize the cost of
the can,
 The radius should be r  3 500 /  cm
 The height should be equal to twice the radius—
namely, the diameter
OPTIMIZATION PROBLEMS
Example 3
Find the point on the parabola
y2 = 2x
that is closest to the point (1, 4).
OPTIMIZATION PROBLEMS
Example 3
The distance between
the point (1, 4) and
the point (x, y) is:
d  ( x  1)  ( y  4)
 However, if (x, y) lies on
the parabola, then x = ½ y2.
 So, the expression for d
becomes:
d  ( 12 y 2  1)2  ( y  4)2
2
2
OPTIMIZATION PROBLEMS
Example 3
Instead of minimizing d, we minimize
its square:
d  f ( y )   y  1   y  4 
2
1
2
2
2
2
 You should convince yourself that the minimum of d
occurs at the same point as the minimum of d2.
 However, d2 is easier to work with.
OPTIMIZATION PROBLEMS
Example 3
Differentiating, we obtain:
f '( y )  2  y  1 y  2( y  4)  y  8
1
2
2
So, f’(y) = 0 when y = 2.
3
OPTIMIZATION PROBLEMS
Example 3
Observe that f’(y) < 0 when y < 2 and f’(y) > 0
when y > 2.
So, by the First Derivative Test for Absolute
Extreme Values, the absolute minimum
occurs when y = 2.
f '( y)  y  8
3
OPTIMIZATION PROBLEMS
The corresponding value of x is:
x = ½ y2 = 2
Thus, the point on y2 = 2x
closest to (1, 4) is (2, 2).
Example 3
OPTIMIZATION PROBLEMS
A man launches his boat
from point A on a bank of
a straight river, 3 km wide,
and wants to reach point B
(8 km downstream on
the opposite bank) as
quickly as possible.
Example 4
OPTIMIZATION PROBLEMS
He could proceed in any
of three ways:
 Row his boat directly across
the river to point C and then run
to B
 Row directly to B
 Row to some point D between
C and B and then run to B
Example 4
OPTIMIZATION PROBLEMS
If he can row 6 km/h and
run 8 km/h, where should he
land to reach B as soon as
possible?
 We assume that the speed of
the water is negligible compared
with the speed at which he rows.
Example 4
OPTIMIZATION PROBLEMS
Example 4
If we let x be the distance from C to D,
then:
 The running distance is:
|DB| = 8 – x
 The Pythagorean Theorem gives
the rowing distance as:
|AD| = x2  9
OPTIMIZATION PROBLEMS
Example 4
distance
We use the equation time=
rate
 Then, the rowing time is: x 2  9 / 6
 The running time is: (8 – x)/8
 So, the total time T as a function
of x is:
T ( x) 
x2  9 8  x

6
8
OPTIMIZATION PROBLEMS
Example 4
The domain of this function T is [0, 8].
 Notice that if x = 0, he rows to C, and if x = 8,
he rows directly to B.
x
1
 The derivative of T is: T '( x) 

6 x2  9 8
OPTIMIZATION PROBLEMS
Example 4
Thus, using the fact that x ≥ 0,
x
1
we have: T '( x)  0 

2
6 x 9 8
 4 x  3 x2  9
 16 x  9( x  9)
2
2
9
 7 x  81  x 
7
2
 The only critical number is: 9 / 7 ≈ 3.4
OPTIMIZATION PROBLEMS
Example 4
To see whether the minimum occurs at
this critical number or at an endpoint of
the domain [0, 8], we evaluate T at all three
points:
T (0)  1.5
7
 9 
T
  1  8  1.33
 7
73
T (8) 
 1.42
6
OPTIMIZATION PROBLEMS
Since the smallest of
these values of T
occurs when x = 9 /
7,
the absolute minimum
value of T must occur
there.
 The figure illustrates
this calculation by
showing the graph
of T.
Example 4
OPTIMIZATION PROBLEMS
Thus, the man should land the
boat at a point
9 / 7 (≈ 3.4 km)
downstream from his starting
point.
Example 4