Chapter 7
Calculus-Based Solutions Procedures
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Nonlinear Optimization
Review of Derivatives
 Models with One Decision Variable
 Unconstrained Models with More Than One
Decision Variable
 Models with Equality Constraints:

 Lagrange


Multipliers
Interpretation of Lagrange Multiplier
Models Involving Inequality Constraints
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Review of 1st Derivatives

Notation:
y
= f(x), dy/dx = f’(x)
f(x) = c
 f(x) = xn
 f(x) = x
 f(x) = x5
 f(x) = 1/x3
 f(x) = c*g(x)
 f(x) = 10*x2
 f(x) = u(x)+v(x)
 f(x) = x2 - 5x

f’(x) = 0
f’(x) = n*x(n-1)
f’(x) = 1*x0 = 1
f’(x) = 5*x4
f(x) = x-3
f’(x) = -3*x-4
f’(x) = c*g’(x)
f’(x) = 20*x
f’(x) = u’(x)+v’(x)
f’(x) = 2x - 5
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Review of 2nd Derivatives

Notation:
y
= f(x), d(f’(x))/dx = d2y/dx2 = f’’(x)
f(x) = -x2
 f(x) = x-3

f’(x) = -2x
f’(x) = -3x-4
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f’’(x) = -2
f’’(x) = 12x-5
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Nonlinear Optimization
Review of Derivatives
 Models with One Decision Variable
 Unconstrained Models with More Than One
Decision Variable
 Models with Equality Constraints:

 Lagrange


Multipliers
Interpretation of Lagrange Multiplier
Models Involving Inequality Constraints
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Models with One Decision Variable

Requires 1st & 2nd derivative tests
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1st & 2nd Derivative Tests

Rule 1 (Necessary Condition):
 df/dx

=0
Rule 2 (Sufficient Condition):
 d2f/dx2
>0
 d2f/dx2 < 0
Minimum
Maximum
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Maximum Example

Rule 1:
= y = -50 + 100x – 5x2
 dy/dx = 100 – 10x = 0, x = 10
 f(x)

Rule 2:
 d2y/dx2

= -10
Therefore, since d2y/dx2 < 0: f(x) has a Maximum at
x=10
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Y = f(x)
Maximum Example – Graph Solution
500
450
400
350
300
250
200
150
100
50
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
x
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Minimum Example

Rule 1:
= y = x2 – 6x + 9
 dy/dx = 2x - 6 = 0, x = 3
 f(x)

Rule 2:
 d2y/dx2

=2
Therefore, since d2y/dx2 > 0: f(x) has a Minimum at
x=3.
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Minimum Example – Graph Solution
400
350
Y = f(x)
300
250
200
150
100
50
0
1
4
7
3
10 13 16 19 22 25 28 31 34 37
x
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Max & Min Example

Rule 1:
= y = x3/3 – x2
 dy/dx = f’(x) = x2 – 2x = 0; x = 0, 2
 f(x)

Rule 2:
= f’’(x) = 2x – 2 = 0
 2(0) – 2 = -2, f’’(x=0) = -2
 d2y/dx2

Therefore, d2y/dx2 < 0: Maximum of f(x) at x=0
 2(2)

– 2 = 2, f’’(x=2) = 2
Therefore, d2y/dx2 > 0: Minimum of f(x) at x=2
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Max & Min Example – Graph Solution
6
4
Y = f(x)
2
0
-2
2
0
1
4
7
10 13 16 19 22 25 28 31 34 37
-4
-6
-8
x
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Example: Cubic Cost Function
Resulting in Quadratic 1st Derivative

Rule 1:



f(x) = C = 10x3 – 200x2 – 30x + 15,000
dC/dx = f’(x)= 30x2 – 400x – 30 = 0
Quadratic Form: ax2 + bx + c
x  b
b^ 2  4ac
2a
x  400 
(400)^ 2  4(30)( 30)
2(30)
x  13.4,.07
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Example: Cubic Cost Function
Resulting in Quadratic 1st Derivative

Rule 2:
= f’’(x) = 60x – 400
 60(13.4) – 400 = 404 > 0
 d2y/dx2

Therefore, d2y/dx2 > 0: Minimum of f(x) at x = 13.4
 60(-.07)

– 400 = -404.2 < 0
Therefore, d2y/dx2 < 0: Maximum of f(x) at x = -.07
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Cost $ (C) = f(x)
Cubic Cost Function – Graph Solution
50000
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
1
4 -.077 10 13 16 13.4
19 22 25 28 31 34 37
x = Units Produced
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Economic Order Quantity – EOQ

Assumptions:





Demand for a particular item is known and constant
Reorder time (time from when the order is placed until the
shipment arrives) is also known
The order is filled all at once, i.e. when the shipment
arrives, it arrives all at once and in the quantity requested
Annual cost of carrying the item in inventory is
proportional to the value of the items in inventory
Ordering cost is fixed and constant, regardless of the size of
the order
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Economic Order Quantity – EOQ

Variable Definitions:
 Let
Q represent the optimal order quantity, or the EOQ
 Ch represent the annual carrying (or holding) cost per
unit of inventory
 Co represent the fixed ordering costs per order
 D represent the number of units demanded annually

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Economic Order Quantity – EOQ

Note: If all the previous assumptions are
satisfied, then the number of units in inventory
would follow the pattern in the graph below:
EOQ Model
Q
Time
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Economic Order Quantity – EOQ

At time = 0 after the initial delivery, the
inventory level would be Q. The inventory
level would then decline, following the straight
line since demand is constant. When the
inventory just reaches zero, the next delivery
would occur (since delivery time is known and
constant) and the inventory would
instantaneously return to Q. This pattern would
repeat throughout the year.
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Economic Order Quantity – EOQ

Under these assumptions:
 Average
Inventory Level = Q/2
 Annual Carrying (or Holding) Cost = (Q/2)*Ch
 The annual ordering cost would be the number of
orders times the ordering cost: (D/Q)* Co
 Total Annual Cost = TC = (Q/2)*Ch+(D/Q)* Co
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Economic Order Quantity – EOQ

To find the Optimal Order Quantity, Q take the first
derivative of TC with respect to Q:


(dTC/dQ) = (Ch/2) – DCoQ-2 = 0
Solving this for Q, we find:

Q* = (2DCo/Ch)^(1/2)


Which is the Optimal Order Quaintly
Checking the second-order conditions (Rule 2 in our
text), we have:

(d2TC/dQ2)= (2DCo/Q3)

Which is always > 0, since all the quantities in the expression are
positive. Therefore, Q* gives a minimum value for total cost (TC)
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Nation’s Healthcare Inc (NHI) has collected historical data on the cost of operating a large
hospital. The operating cost turns out to be a nonlinear function of the number of patient
days per year, approximated by the function:
C  4,700,00  .00013x 2
where C is the total annual cost and x is the number of patient days per year.
a. Write the equation for cost per patient day.
b. Find the value of x (patient days) which minimizes cost per patient day.
c. Find the minimum cost per patient day.
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Orion Outfitters is trying to price a new pair of ski goggles. They have estimates of the
relationship between price and the number of units sold as below:
p  50  .05 x
where p is the price and x is the number of units sold.
a. Write the equation for total revenue.
b. Find the number of units to sell in order to maximize revenue.
c. Find the revenue-maximizing price.
d. Find the maximum revenue.
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SouthStar Inc. (SSI) produces lawn tractors at a single
factory. Based on a number of years of data, SSI has
estimated a nonlinear cost function for the factory as below:
C  100,000  1500 x  .2 x 2
where C is the total annual cost in dollars and x is the number of units produced in a year.
a. Find the number of units to produce in order to minimize cost per unit.
b. What is the minimum cost per unit?
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Nonlinear optimization, one variable, restricted interval
Find the minimum for the cost function:
C  2 x 3  12 x 2  100
where x is the production level in thousands of units and C is the total cost in
millions of dollars. Suppose that, due to other factors, the production level
must be no lower than 1 thousand units and no more than 10 thousand units.
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Nonlinear optimization, one variable, restricted interval
Consider the cost function:
C  2 x 3  6 x 2  10
where x is the production level in thousands of units and C is the total cost in millions of
dollars. Find the production level which yields the minimum cost. Assume that production
must be no lower than 1 thousand units and no higher than 5 thousand units.
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Restricted Interval Problems

Step 1:


Step 2:


Find all the points that satisfy rules 1 & 2. These are
candidates for yielding the optimal solution to the problem.
If the optimal solution is restricted to a specified interval,
evaluate the function at the end points of the interval.
Step 3:

Compare the values of the function at all the points found
in steps 1 and 2. The largest of these is the global maximum
solution; the smallest is the global minimum solution.
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Nonlinear Optimization
Review of Derivatives
 Models with One Decision Variable
 Unconstrained Models with More Than One
Decision Variable
 Models with Equality Constraints:

 Lagrange


Multipliers
Interpretation of Lagrange Multiplier
Models Involving Inequality Constraints
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General Form – Relative Min. and Max
Relative Maximum @ (-1,0,e)
Relative Minimum @ (2,- 1,-3)
z
z
y
x
y
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General Form – Saddle Point
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Unconstrained Models with More Than
One Decision Variable

Requires partial derivatives
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Example 1st Partial Derivatives

If z = 3x2y3
 ∂z/∂x
= 6xy3
 ∂z/∂y = 9y2x2

If z = 5x3 – 3x2y2 + 7y5
 ∂z/∂x
= 15x2 – 6xy2
 ∂z/∂y = -6x2y + 35y4
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2nd Partial Derivatives

2nd Partials Notation
= ∂2z/∂x2
 (∂/dy)*(∂z/∂y) = ∂2z/∂y2
 (∂/∂x)*(∂z/∂x)

Mixed Partials Notation
= ∂2z/(∂x∂y)
 (∂/∂y)*(∂z/∂x) = ∂2z/(∂y∂x)
 (∂/∂x)*(∂z/∂y)
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Example 2nd Partial Derivatives

If z = 7x3 + 9xy2 + 2y5
 ∂z/∂x
= 21x2 + 9y2
 ∂z/∂y = 18xy + 10y4
 ∂2z/(∂y∂x) = 18y
 ∂2z/(∂x∂y) = 18y
 ∂2z/∂x2 = 42x
 ∂2z/∂y2 = 18x + 40y3
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Partial Derivative Tests

Rule 3 (Necessary Condition):
 ∂f/∂x1

= 0, ∂f/∂x2 = 0, Solve Simultaneously
Rule 4 (Sufficient Condition):
 If ∂2f/∂x12
>0
 And (∂2f/∂x12)*(∂2f/∂x22) – (∂2f/(∂x1∂x2))2 > 0

Then Minimum
 If ∂2f/∂x12
<0
 And (∂2f/∂x12)*(∂2f/∂x22) – (∂2f/(∂x1∂x2))2 > 0

Then Maximum
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Partial Derivative Tests

Rule 4, continued:
 If

 If

(∂2f/∂x12)*(∂2f/∂x22) – (∂2f/(∂x1∂x2))2 < 0
Then Saddle Point – Neither Maximum nor Minimum
(∂2f/∂x12)*(∂2f/∂x22) – (∂2f/(∂x1∂x2))2 = 0
Then no conclusion
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Optimization with two variables (bivariate optimization), no constraints
A company is trying to construct an advertising plan. They can choose between TV advertising and
radio advertising. From previous experience they have found that the following equation
approximates the relationship between sales and advertising expenditures:
f ( x, y)  50,000 x  40,000 y  10 x 2  20 y 2  10 xy
Where f(x,y) is unit sales, x is dollars spent on TV ads. And y is dollars spent on radio ads. Find the
advertising plan which will result in maximum sales.
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A manufacturer sells two products. The demand functions for these two products are as given below:
q1  150  2 p1  p 2
q 2  200  p1  3 p 2
where q1 is the number of units of product 1 sold, q2 is the number of units of product 2 sold, p1 is the
price of product 1 in dollars and p2 is the price of product 2 in dollars. Find the prices that the
manufacturer should charge in order to maximize revenue.
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A service company sells two products. Below is given the profit function of the company as a function of
the number of units of each product produced.
f ( x, y)  64 x  2 x 2  4 xy  4 y 2  32 y  14
where f(x,y) is profit, x is the number of units of product one sold, and y is the number of units of product
two sold. Find the number of units of each product that should be sold in order to maximize profit.
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Nonlinear Optimization
Review of Derivatives
 Models with One Decision Variable
 Unconstrained Models with More Than One
Decision Variable
 Models with Equality Constraints:

 Lagrange
Multipliers
 Interpretation

of Lagrange Multiplier
Models Involving Inequality Constraints
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Lagrange Multipliers

Nonlinear Optimization with an equality
constraint
 Max
or Min f(x1, x2)
 ST: g(x1, x2) = b

Form the Lagrangian Function:
L=
f(x1, x2) + λ[g(x1, x2) – b]
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Lagrange Multipliers

Rule 6 (Necessary Condition):
 Optimization
of an equality constrained function,
1st order conditions:
∂L/∂x1 = 0
 ∂L/∂x2 = 0
 ∂L/∂λ = 0

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Lagrange Multipliers

Rule 7 (Sufficient Condition):
rule 6 is satisfied at a point (x*1, x*2, λ*) apply
conditions (a) and (b) of rule 4 to the Lagrangian
function with λ fixed at a value of λ* to determine
if the point (x*1, x*2) is a local maximum or a local
minimum.
 If
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Interpretation of Lagrange Multipliers

The value of the Lagrange multiplier
associated with the general model above is the
negative of the rate of change of the objective
function with respect to a change in b. More
formally, it is negative of the partial derivative
of f(x1, x2) with respect to b; that is,
λ
= - ∂f/∂b or
 ∂f/∂b = - λ
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A company has a requirement to produce 34 units of a new product. The order can be filled by either
product 1 or product 2 or a combination of the two. The company’s cost function is:
C  6Q1  10Q2  Q1Q2  30
2
2
a. How many units of each product should be produced in order to
minimize total cost?
b. What is the minimum cost?
c. What would be the effect on cost of a one unit increase in the total
production requirement?
d. Now solve this problem on EXCEL.
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Nonlinear Optimization, two variables with a constraint
Itech Cycle Company (ICC) has an order to produce 200 bicycles. ICC produces this particular bicycle at two plants. The cost
function for production at these two plants is:
Cost : f ( x1 , x 2 )  2 x1  x1 x 2  x 2  200
2
2
Where f(x1,x2) is the production cost in dollars, x1 is the number of bicycles produced at plant 1 and x2 is the number of bicycles
produced at plant 2. The company wants to split the production between the two plants in such a way as to minimize production
cost.
a. How many bicycles should ICC produce at each plant in order to meet the order at minimum cost?
b. What is the minimum cost?
c. What would be the effect on cost of a one unit increase in the total production requirement?
d. Now solve this problem on EXCEL.
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Nonlinear Optimization
Review of Derivatives
 Models with One Decision Variable
 Unconstrained Models with More Than One
Decision Variable
 Models with Equality Constraints:

 Lagrange


Multipliers
Interpretation of Lagrange Multiplier
Models Involving Inequality Constraints
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Models Involving Inequality Constraints

Step 1:


Assume the constraint is not binding, and apply the procedures of
“Unconstrained Models with More Than One Decision Variable” to
find the global maximum of the function, if it exists. (Functions that go
to infinity do not have a global maximum). If this global maximum
satisfies the constraint, stop. This is the global maximum for the
inequality-constrained problem. If not, the constraint may be binding at
the optimum. Record the value of any local maximum that satisfies the
inequality constraint, and go on to Step 2.
Step 2:

Assume the constraint is binding, and apply the procedures of “Models
with Equality Constraints” to find all the local maxima of the resulting
equality-constrained problem. Compare these values with any feasible
local maxima found in Step 1. The largest of these is the global
maximum.
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