Question 1

The optimal solution to the nonlinear model : Minimize: $f(x)=10 X_{2}+40 X+1$, subject to $X^{3} 3$, is given by $\mathrm{X}$ equal to

Accepted Answer

A)  -29 
B)  -39 
C)  0 
D)  1 
A)  -29 
B)  -39 
C)  0 
D)  1

Question 2

If the first partial derivative of a Lagrangian function of two decision variables are equal to zero at a point, then the point is

Accepted Answer

A) a saddle point maximum if the mixed partial is negative B) a local maximum if the second partials are negative and $\mathrm{D}>0$ C) a global maximum if the second and mixed partials are all positive D) a global minimum if the second partials are positive and $\mathrm{D}<0$ A) a saddle point maximum if the mixed partial is negative B) a local maximum if the second partials are negative and $\mathrm{D}>0$ C) a global maximum if the second and mixed partials are all positive D) a global minimum if the second partials are positive and $\mathrm{D}<0$

Question 3

The local minimum of a convex function will also be the global minimum.

Accepted Answer

A) True 
 B)False

Question 4

A two-variable problem with an inequality constraint may, in some instances, be solvable by ignoring the constraint because it may not be binding.

Accepted Answer

A) True 
 B)False

Question 5

The optimal solution to the nonlinear model : Minimize: $f(x)=3 X_{3}-20 X_{2}+150$, subject to $X^{3} 0$ and $X$ $\pounds 5$, is given by $X$ equal to

Accepted Answer

A)  0 
B)  4.4444 
C)  2.828 
D)  5 
A)  0 
B)  4.4444 
C)  2.828 
D)  5

Question 6

As the number of turning points of a continuous function $\mathrm{f}(\mathrm{x})$ increases, the number of local maximums and minimums will increase.

Accepted Answer

A) True 
 B)False

Question 7

the function $\mathrm{f}$.
$D=\frac{\partial^{2} f}{\partial X_{1}^{2}} \cdot \frac{\partial^{2} f}{\partial X_{2}^{2}}-\left(\frac{\partial^{2} f}{\partial X_{1} \partial X_{2}}\right)^{2}>0$

Accepted Answer

A) True 
 B)False

Question 8

Necessary and sufficient conditions for the existence of a local minimum in a single- variable, unconstrained, nonlinear optimization problem are that the second derivative be negative at a point and at the same point the slope of the function be 0 .

Accepted Answer

A) True 
 B)False

Question 9

A necessary condition for a one-variable decision problem with an objective function, which is continuous, is that the first derivative of the function be 0 at a local maximum or minimum point.

Accepted Answer

A) True 
 B)False

Question 10

The local minimum for the function $f(X)=8 X_{2}-4 X-3$ occurs at $X$ equal to

Accepted Answer

A)  2 
B)  5 
C)  0.25 
D)  4 
A)  2 
B)  5 
C)  0.25 
D)  4

Question 11

In an unconstrained two-variable problem with a quadratic objective function, the saddle point is where one variable reaches a local maximum and the other variable reaches a local minimum.

Accepted Answer

A) True 
 B)False

Question 12

Which of these factors is essential in deciding when to use linear approximation of nonlinear problems?

Accepted Answer

A)  Feasibility 
B)  Experience 
C)  Efficiency 
D)  Performance 
A)  Feasibility 
B)  Experience 
C)  Efficiency 
D)  Performance

Question 13

If the slope of the function $f(x)=4 X_{2}-24 X+32$ is 0 at $X=3$, then it follows that $X=3$ is a

Accepted Answer

A)  local maximum 
B)  local minimum 
C)  global maximum 
D)  saddle point 
A)  local maximum 
B)  local minimum 
C)  global maximum 
D)  saddle point

Question 14

The optimal solution to the nonlinear model : Maximize: $f(x)=5 X_{2}-30 X+50$, subject to $X^{3} 0$ and $X \pounds$ 4 , is given by $X$ equal to

Accepted Answer

A)  4 
B)  0 
C)  2 
D)  1 
A)  4 
B)  0 
C)  2 
D)  1

Question 15

A quadratic function $\mathrm{f}(\mathrm{x})$ has one local and global maximum or one local and global minimum.

Accepted Answer

A) True 
 B)False

Question 16

In a small, specialty door-locks business, demand is given by $\mathrm{d}=100-2 \mathrm{p}$, where $\mathrm{d}$ is the monthly demand in units and $\mathrm{p}$ is the price per unit. Identify the expression that correctly captures

Accepted Answer

A)  total sales revenue 
B)  It is easier to plot 
C)  Local maximum is less than global maximum 
D)  Local minimum is equal to global minimum 
E)  Local minimum is greater than global minimum 
A)  total sales revenue 
B)  It is easier to plot 
C)  Local maximum is less than global maximum 
D)  Local minimum is equal to global minimum 
E)  Local minimum is greater than global minimum

Question 17

The value of the function $f(x)=4 X_{3}+2 X_{2}-4 X-10$ at its local maximum is

Accepted Answer

A)  -8 
B)  8 
C)  1 
D)  -1 
A)  -8 
B)  8 
C)  1 
D)  -1

Question 18

Which of the following is true for a strictly convex function?

Accepted Answer

A)  It is easier to plot 
B)  Local maximum is less than global maximum 
C)  Local minimum is equal to global minimum 
D)  Local minimum is greater than global minimum 
A)  It is easier to plot 
B)  Local maximum is less than global maximum 
C)  Local minimum is equal to global minimum 
D)  Local minimum is greater than global minimum

Question 19

Partial derivatives are used to find extreme values of decision variables in a nonlinear model with 2 or more variables.

Accepted Answer

A) True 
 B)False

Question 20

The optimal solution to the nonlinear model : Maximize: $f(x)=3 X_{3}-15 X_{2}+24 X$, subject to $X \pounds 2$, is given by $\mathrm{X}$ equal to

Accepted Answer

A)  2 
B)  1.5 
C)  1.333 
D)  1.867 
A)  2 
B)  1.5 
C)  1.333 
D)  1.867

The optimal solution to the nonlinear model : Minimize: $f(x)=10 X_{2}+40 X+1$ , subject to $X^{3} 3$ , is given by $\mathrm{X}$ equal to

If the first partial derivative of a Lagrangian function of two decision variables are equal to zero at a point, then the point is

The local minimum of a convex function will also be the global minimum.

A two-variable problem with an inequality constraint may, in some instances, be solvable by ignoring the constraint because it may not be binding.

The optimal solution to the nonlinear model : Minimize: $f(x)=3 X_{3}-20 X_{2}+150$ , subject to $X^{3} 0$ and $X$ $\pounds 5$ , is given by $X$ equal to

As the number of turning points of a continuous function $\mathrm{f}(\mathrm{x})$ increases, the number of local maximums and minimums will increase.

the function $\mathrm{f}$ . $D=\frac{\partial^{2} f}{\partial X_{1}^{2}} \cdot \frac{\partial^{2} f}{\partial X_{2}^{2}}-\left(\frac{\partial^{2} f}{\partial X_{1} \partial X_{2}}\right)^{2}>0$

Necessary and sufficient conditions for the existence of a local minimum in a single- variable, unconstrained, nonlinear optimization problem are that the second derivative be negative at a point and at the same point the slope of the function be 0 .

A necessary condition for a one-variable decision problem with an objective function, which is continuous, is that the first derivative of the function be 0 at a local maximum or minimum point.

The local minimum for the function $f(X)=8 X_{2}-4 X-3$ occurs at $X$ equal to

In an unconstrained two-variable problem with a quadratic objective function, the saddle point is where one variable reaches a local maximum and the other variable reaches a local minimum.

Which of these factors is essential in deciding when to use linear approximation of nonlinear problems?

If the slope of the function $f(x)=4 X_{2}-24 X+32$ is 0 at $X=3$ , then it follows that $X=3$ is a

The optimal solution to the nonlinear model : Maximize: $f(x)=5 X_{2}-30 X+50$ , subject to $X^{3} 0$ and $X \pounds$ 4 , is given by $X$ equal to

A quadratic function $\mathrm{f}(\mathrm{x})$ has one local and global maximum or one local and global minimum.

In a small, specialty door-locks business, demand is given by $\mathrm{d}=100-2 \mathrm{p}$ , where $\mathrm{d}$ is the monthly demand in units and $\mathrm{p}$ is the price per unit. Identify the expression that correctly captures

The value of the function $f(x)=4 X_{3}+2 X_{2}-4 X-10$ at its local maximum is

Which of the following is true for a strictly convex function?

Partial derivatives are used to find extreme values of decision variables in a nonlinear model with 2 or more variables.

The optimal solution to the nonlinear model : Maximize: $f(x)=3 X_{3}-15 X_{2}+24 X$ , subject to $X \pounds 2$ , is given by $\mathrm{X}$ equal to

Introduction to Management Science, Modeling, and Excel Spreadsheets

Forecasting

Linear Programming: Basic Concepts and Graphical Solutions

Linear Programming: Applications and Solutions

Linear Programming: Sensitivity Analysis, Duality, and Specialized Models

Transportation, Assignment, and Transshipment Problems

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Network Optimization Models

Multi-Criteria Models

Decision Theory

Markov Analysis

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Exam 9: Nonlinear Optimization Models

The optimal solution to the nonlinear model : Minimize: f(x)=10X2+40X+1f(x)=10 X_{2}+40 X+1f(x)=10X2​+40X+1 , subject to X33X^{3} 3X33 , is given by X\mathrm{X}X equal to

If the first partial derivative of a Lagrangian function of two decision variables are equal to zero at a point, then the point is

The local minimum of a convex function will also be the global minimum.

A two-variable problem with an inequality constraint may, in some instances, be solvable by ignoring the constraint because it may not be binding.

The optimal solution to the nonlinear model : Minimize: f(x)=3X3−20X2+150f(x)=3 X_{3}-20 X_{2}+150f(x)=3X3​−20X2​+150 , subject to X30X^{3} 0X30 and XXX £5\pounds 5£5 , is given by XXX equal to

As the number of turning points of a continuous function f(x)\mathrm{f}(\mathrm{x})f(x) increases, the number of local maximums and minimums will increase.

Necessary and sufficient conditions for the existence of a local minimum in a single- variable, unconstrained, nonlinear optimization problem are that the second derivative be negative at a point and at the same point the slope of the function be 0 .

A necessary condition for a one-variable decision problem with an objective function, which is continuous, is that the first derivative of the function be 0 at a local maximum or minimum point.

The local minimum for the function f(X)=8X2−4X−3f(X)=8 X_{2}-4 X-3f(X)=8X2​−4X−3 occurs at XXX equal to

In an unconstrained two-variable problem with a quadratic objective function, the saddle point is where one variable reaches a local maximum and the other variable reaches a local minimum.

Which of these factors is essential in deciding when to use linear approximation of nonlinear problems?

If the slope of the function f(x)=4X2−24X+32f(x)=4 X_{2}-24 X+32f(x)=4X2​−24X+32 is 0 at X=3X=3X=3 , then it follows that X=3X=3X=3 is a

The optimal solution to the nonlinear model : Maximize: f(x)=5X2−30X+50f(x)=5 X_{2}-30 X+50f(x)=5X2​−30X+50 , subject to X30X^{3} 0X30 and X£X \poundsX£ 4 , is given by XXX equal to

A quadratic function f(x)\mathrm{f}(\mathrm{x})f(x) has one local and global maximum or one local and global minimum.

In a small, specialty door-locks business, demand is given by d=100−2p\mathrm{d}=100-2 \mathrm{p}d=100−2p , where d\mathrm{d}d is the monthly demand in units and p\mathrm{p}p is the price per unit. Identify the expression that correctly captures

The value of the function f(x)=4X3+2X2−4X−10f(x)=4 X_{3}+2 X_{2}-4 X-10f(x)=4X3​+2X2​−4X−10 at its local maximum is

Which of the following is true for a strictly convex function?

Partial derivatives are used to find extreme values of decision variables in a nonlinear model with 2 or more variables.

The optimal solution to the nonlinear model : Maximize: f(x)=3X3−15X2+24Xf(x)=3 X_{3}-15 X_{2}+24 Xf(x)=3X3​−15X2​+24X , subject to X£2X \pounds 2X£2 , is given by X\mathrm{X}X equal to

Introduction to Management Science, Modeling, and Excel Spreadsheets

Forecasting

Linear Programming: Basic Concepts and Graphical Solutions

Linear Programming: Applications and Solutions

Linear Programming: Sensitivity Analysis, Duality, and Specialized Models

Transportation, Assignment, and Transshipment Problems

Integer Programming

Network Optimization Models

Multi-Criteria Models

Decision Theory

Markov Analysis

Waiting Line Models

Simulation Cdrom Modules

Filters

The optimal solution to the nonlinear model : Minimize: $f(x)=10 X_{2}+40 X+1$ , subject to $X^{3} 3$ , is given by $\mathrm{X}$ equal to

The optimal solution to the nonlinear model : Minimize: $f(x)=3 X_{3}-20 X_{2}+150$ , subject to $X^{3} 0$ and $X$ $\pounds 5$ , is given by $X$ equal to

As the number of turning points of a continuous function $\mathrm{f}(\mathrm{x})$ increases, the number of local maximums and minimums will increase.

The local minimum for the function $f(X)=8 X_{2}-4 X-3$ occurs at $X$ equal to

If the slope of the function $f(x)=4 X_{2}-24 X+32$ is 0 at $X=3$ , then it follows that $X=3$ is a

The optimal solution to the nonlinear model : Maximize: $f(x)=5 X_{2}-30 X+50$ , subject to $X^{3} 0$ and $X \pounds$ 4 , is given by $X$ equal to

A quadratic function $\mathrm{f}(\mathrm{x})$ has one local and global maximum or one local and global minimum.

In a small, specialty door-locks business, demand is given by $\mathrm{d}=100-2 \mathrm{p}$ , where $\mathrm{d}$ is the monthly demand in units and $\mathrm{p}$ is the price per unit. Identify the expression that correctly captures

The value of the function $f(x)=4 X_{3}+2 X_{2}-4 X-10$ at its local maximum is

The optimal solution to the nonlinear model : Maximize: $f(x)=3 X_{3}-15 X_{2}+24 X$ , subject to $X \pounds 2$ , is given by $\mathrm{X}$ equal to