15.4 Constrained Maxima and Minima

Transcription

1 15.4 Constrained Maxima and Minima Question 1: Ho do ou find the relative extrema of a surface hen the values of the variables are constrained? Question : Ho do ou model an optimization problem ith several variables subject to a constraint? In an earlier chapter, ou learned ho to optimize functions developed from applications. These problems ere solved b forming a table. The table helped ou to rite the function to be maximized or minimized. The function could be optimized using the first and second derivatives. Man of these objective functions ma be ritten as multivariable functions ith the variables constrained. In this section, ou ll learn ho to appl Lagrange multipliers to solve these tpes of optimization problems. 1

2 Question 1: Ho do ou find the relative extrema of a surface hen the values of the variables are constrained? In Section 1.3, ou learned ho to find the relative maximum or relative minimum on a surface. No e ill find relative maximum and minimum along a specific portion of the surface defined b a constraint. f x, x 3x 15 Suppose e ant to find the relative maximum of the surface subject to x 4. This means e ant to find points on the surface that are higher than the other nearb points here the plane x 4 intersects the surface. x 4 z x x 3 15 Figure 1 - The plane intersects the surface along the path traced b the black curve. The relative maximum is found b defining a ne function called F in terms of the surface and the constraint. If there are an relative extrema along the path, the ill occur at critical points of this ne function.

3 Lagrange Multiplier Method Suppose a surface is given b z f x, and that the surface is subject to a constraint g x, 0. The relative extrema of the surface on the portion corresponding to the constraint is found b folloing the steps belo. 1. Define the function F x,, f x, g x,.. Set all first partial derivatives of F x,, equal to zero, F x,, 0, F x,, 0, F x,, 0 x 3. Find the critical points of F x,, b solving the resulting sstem of equations. An relative extrema along the portion of the surface defined b the constraint is at these critical points. The variable is called a Lagrange multiplier. We have a surface f, but not a constraint matching g. To get g, e need to rite the constraint so that zero appears on one side. Subtract 4 from both sides of the constraint to give x 40 g x, The other side of the constraint is g. Once the constraint is in the proper format, e appl the steps above to find the relative maximum. 3

4 Example 1 Maximize the Constrained Function Maximize f x, x 3 15 subject to the constraint g x, x 4. here g x, 0 Solution Define the function F using the Lagrange multiplier, F x,, x 3 15 x4 f x, g x, x 3 15 x4 The partial derivatives are F x,, 4x x F x,, 6 F x,, x 4 The critical points are found b solving the sstem of equations, 4x x 40 The first and second equation are eas to solve for. Solve each for and them set them equal to eliminate, 4x 6 x 6 4 Solve for x b dividing both sides b 4 After solving for one of the variables (in this case x), substitute the resulting equation into the third equation, 4

5 Replace x ith 6 4 Combine like terms Add 4 to both sides Multipl both sides b 4 10 and simplif. 6 Since x, the corresponding x value is x The critical point is.4,1.6, but e don t kno hether this point is a relative maximum or relative minimum. We can decide this b examining a table of values for the function. x f x, x In each column, the x and values satisf the constraint x 4 0. The function is largest at the critical point so it must correspond to a relative maximum. If a graph is available, e can also use it to identif that the critical point is a maximum. (.4, 1.6) 5

6 In practice, the most difficult part of the Lagrange Multiplier Method is solving the sstem of equations. A good strateg is to solve F x,, 0 and x F x,, 0 for. Setting these equations equal to each other eliminates from the sstem. Then e can use the resulting relationship in F x,, 0 to find the critical point. Example Minimize the Constrained Function f x, x subject to x 1. Minimize Solution The constraint must be ritten in the form move all of the terms to the left hand side, e have g x, 0. If e x 10 g x, We ll find the critical points of F x,, x x1 f x, g x, using the partial derivatives F x,, x x F x,, F x,, x1 Set each partial derivative equal to zero and solve the resulting sstem for x and. x 0 x 0 x Solve each equation for and then set equal to each other 6

7 x x 10 x 4x 10 Substitute x into the third equation in the sstem and solve for x 3x 10 3x 1 x 1 3 Using x point is 1, the corresponding value is or 3. The critical,. To determine hether this is a maximum or a minimum, let s look at a table of values near the critical point. 3 x , f x x The x and values in the table must be picked in the table so that the satisf the constraint. In this case, values of are picked near 3 and then the constraint is used to find the corresponding value. The loest 1 value of the function occurs at the critical point so 3 is the minimum 1 value of the function. It occurs at,., f x x 3 3 Portion of surface corresponding to constraint,

8 Question : Ho do ou model an optimization problem ith several variables subject to a constraint? In Chapter 1 e developed objective functions for optimization problems. In those problems, e utilized constraints in the problems to help us identif the function to be optimized. In this question, e model the optimization problems ith several variables and a constraint. This allos us to appl the Lagrange Multiplier Method to find the maximum or minimum of the objective function. Example 3 Minimize Cost of Materials A farmer is fencing a rectangular area for to equall sized pens. These pens share a divider that ill be constructed from chicken ire costing $0.60 per foot. The rest of the pen ill be built from fencing costing $1.10 per foot. All other sides cost $1.10 per foot Divider cost $0.60 per foot If the pens should enclose a total of 400 square feet, hat overall dimensions should the pens have to minimize fencing costs? Solution Identif the variables in the problem. Since the problem asks us to find the overall dimensions of the pens, define the idth of the enclosure as and the length as l. These are label on the diagram. 8

9 l All other sides cost $1.10 per foot Divider cost $0.60 per foot Since e kno the cost per foot for each side, e ill use that information to rite don the cost in terms of l and,, C l l l cost of left and right sides cost of top and bottom sides cost of divider This simplifies to C l,.80l.0 The pens must enclose 400 square feet. In terms of the variables, the area is l so e have the constraint l 400 In the proper format for the Lagrange Multiplier Method, this becomes l The Lagrange function is,, F l l l The partial derivatives of this function are 9

10 F l,,.8 l F l,,.l F l,, l400 We ll find the critical points b solving the sstem of equations,.8 0. l 0 l The first to equations are solved for to ield l 0. l Set the equations equal to each other to eliminate,.8. l You can solve for either variables, but in this case e ll solve for l to give l..8 Substitute this expression into the third equation and solve for : 10

11 Replace l ith..8 Isolate Square root both sides of the equation.56 Onl the positive value makes sense as a dimension of the enclosure. The corresponding length is found from the constraint l 400. Solving for l and sustituting the idth gives 400 l 400 l 400 l A table of values formed from the constraint and objective function verifies that the critical point is a minimum. 400 l C l,.80l The enclosure that costs the least is approximatel.56 feet long and feet ide and costs $99.8. Example 4 Maximize Volume Most airlines charge to check baggage on flights. To avoid these charges, passengers pack as much as possible into their carr-on bags. Hoever, airlines also limit the size of these bags. American Airlines 11

12 limits the linear dimensions (defined as the sum of the length, idth and height) to 45 inches. A manufacturer ishes to produce a carr-on bag hose linear dimensions are 45 inches. The shape of the bag is a rectangular solid, like the one pictured, belo hose ends are squares. What are the dimensions of the bag if its volume is to be as large as possible? Solution As in Example 3, start b labeling the dimensions of the bag. Since the ends are square, the idth and height must be the same. We ll use the letter for both of these dimensions and l for the length of the bag. l In terms of these variables, the volume to be maximized is, l V l Hoever, e must make sure the the sum of the length, idth and height is 45. This gives the constraint 1

13 l 45 Subtract 45 from each side to put the constraint in the proper format, l 45 0 We ant to maximize V l, l subject to the constraint l45 0. The strateg in the Lagrange Multiplier Method indicates that maximum of the objective function is at the critical point of,, 45 F l l l The partial derivatives are F l,, l F l,, l F l,, l45 The critical points are the solution to the sstem of equations, 0 l 0 l 450 Solve each of the first to equations for : 0 l 0 l Set these equation equal to each other to find a relationship for the variables, 13

14 l l 0 l 0 0 or l Add l to both sides Factor from each term Set each factor to zero and solve for A idth of zero is certainl not going to maximize the volume. We ll ignore the critical point corresponding to 0. No let l in the third equation to find the other critical point, ll450 3l 450 3l 45 l 15 Since the constraint indicates that l 45, the corresponding ith must be Let s look at a table to see if this critical point is a maximum. l , l V l The largest volume is 3375 cubic inches and occurs hen the lenghth, idth and height are all 15 inches. 14

15 Example 5 Maximize Production Production in the United States from 1899 through 19 is described b the Cobb-Douglas production formula, , 1.01 Q C L C L here C is the number of units of capital and L is the number of unit of labor. Maximize production if the total amount of capital and labor is 100 units. Solution In this problem, e ant to maximize Q C, L 1.01C L subject to CL 100. Before e can appl the Lagrange Multiplier Method, e need to rerite the constraint as C L100 0 We use this constraint to define the function ,, F C L C L C L The partial derivatives are F C, L, C L C F C, L, C L L F C, L, LC100 The solution to the sstem L C100 0 gives the critical points of F. Solve the first to equation for to ield C L C L 15

16 C L C L C L C L Eliminate b setting the to equations equal to each other, C L C L A fe steps of algebra allos us to rite L in terms of C: C L 3C L L 3CL L 3C Divide each side b Multipl each side b Multipl each side b 0.75 C b adding exponents 0.5 L b adding exponents Use this expression in the third equation to eliminate L: 3CC C Replace L ith 3C Simplif and solve for C 4C 100 C 5 Since CL 100, the corresponding value of L is 75. C L , 1.01 Q C L C L Based on the table, the critical point at C 5 and L 75 leads to the highest amount of production at approximatel units. 16