Applications of Optimization Applications of Calculus Fall 2016
- Slides: 29
Applications of Optimization Applications of Calculus - Fall 2016 • Neal Sherman • Caitlyn Small
Optimization • Unconstrained optimization ▫ Single Variable: �First derivative test �Second derivative test ▫ Multi-variable: �Gradient �Hessian Matrix • Constrained optimization ▫ Lagrangian Method
First Derivative Test for Local Extrema A point c is a critical point for if and only if or does not exist. is a maximum if for and for is a minimum if for and for
Second Derivative Test for Local Extrema • If and then f has a local maximum at • If and then f has a local minimum at
Unconstrained Optimization • Only concern is the objective function you are trying to optimize. ▫ Maximize a desired quantity (ex – profit) ▫ Minimize an undesired quantity (ex – cost) Maximizing is equivalent to minimizing
Example: Single Variable without constraints • A company has determined that its total revenue (in dollars) for a product can be modeled by where x is the number of units produced (and sold). What product level will yield a maximum revenue?
Multivariable "Problem on Optimization without Constraint - Leading Lesson. " Problem on Optimization without Constraint - Leading Lesson. N. p. , n. d. Web. 03 Dec. 2016.
How do we confirm that this point is a minimum? A Hessian Matrix is equivalent to a second derivative test for single variable optimization. • If the Hessian Matrix at v* is positive definite, then is a local minimum. • If the Hessian Matrix at v* is negative definite, then is a local maximum. • If the Hessian Matrix at v* is an indefinite matrix, then neither a local maximum or minimum.
A 2 x 2 Hessian Matrix is positive definite if A matrix is positive definite if it is a symmetric matrix with all positive eigenvalues.
Constrained Optimization • Looking to optimize the objective function, but must stay within the bounds of other restrictions. ▫ These restrictions are called constraints. • Example: Buying certain combination of foods that maximizes nutrition while staying within a budget.
Lagrangian Optimization • Absorbs the constraint into the objective function to transform a constrained optimization problem into an unconstrained optimization. • Given an objective function and a constraint the maximum or minimum of the objective function occurs at a critical value of
A manufacturer’s production is modeled by the Cobb-Douglas function below where x represents the units of labor and y represents the units of capital. Each labor unit costs $150 and each capital unit costs $250. The total expenses for labor and capital cannot exceed $50, 000. Find the maximum production level. Larson, Ron, Bruce H. Edwards, and David C. Falvo. "Section 7. 6 Lagrange Multipliers. " Calculus: An Applied Approach. Boston: Houghton Mifflin, 2006. 50411. Print.
λ ≈0. 334 λ is called the marginal productivity of money For each additional dollar spent on production, 0. 334 more units can be produced. What if $70, 000 was available instead for labor and capital?
Linear Programming : Fuel. Pro • Two grades of petroleum stock (Stock A and Stock B) are used to produce two grades of fuel (as shown in table) • The facility can produce at most 8 gallons of premium fuel each hour. • The company makes a profit of $0. 40 per gallon of premium fuel and $0. 30 per gallon of regular fuel.
Although a linear programming model of this nature can be investigated as early as Algebra I, we will investigate how software, such as Maple, can be used for both linear and non-linear models. From LPSolve, we know the optimal solution is at (4, 10).
The Simplex Algorithm is an alternate method for finding solutions using a graphical approach: • Begin at an extreme point on the boundary of the feasible region, and moves to an adjacent extreme point until the optimal solution is found. • Transform inequality constraints into equality constraints through the introduction of slack variables. • These slack variables represent the difference between the larger and smaller sides of the constraint inequalities, and are constrained to be non-negative, so the LP problem in Figure 3 is replaced by:
Using the Simplex Algorithm, our problem is now:
Non-Linear Programming: Con. Pro • Pipe production is a function of material and labor (denoted by x 1 and x 2, respectively). • Each unit of produced pipe generates $1400 in revenue but costs $350 in materials and $200 in labor to produce. • The company has $40, 000 of available funds to spend, and the ratio of labor units to material units must be at least one third to ensure adequate labor.
• Production is given by • This yields a profit function: • And when assembled with the constraints, the nonlinear programming problem becomes:
Lagrange Method to find optimization • Corresponding Lagrangian Function: Note:
At the optimal solution, the gradient of the Lagrangian function, defined below, will equal zero, as below: complementary slackness equations (Analogous to the slack variables used in the simplex method) must also equal zero.
Resources
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