Chapter 7 page 1 MATH 105 FUNCTIONS OF SEVERAL VARIABLES

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1 Chapter 7 page 1 MATH 105 FUNCTIONS OF SEVERAL VARIABLES Evaluate each function at the indicated point. 1. f(x,y) = x 2 xy + y 3 a) f(2,1) = b) f(1, 2) = 2. g(x,y,z) = 2x y + 5z a) g(2, 0, 1) = b) g(3, -2, 4) = 3. F(t,w) = 4w 2 t 2 F(3, -1) = 4. f ( x,y) = xe y + ln x f (e 2,ln2) = 5. A population that grows exponentially satisfies P(A,k,t) = Ae kt, where P is the population at time t, A is the initial population (when t=0) and k is the relative (per capita) growth rate. The population of a certain county is currently 5 million people and is growing at the rate of 3% per year. What will be the population in 7 years? P(,, ) =

2 Chapter 7 page 2 6. The Cobb-Douglas production function states that x units of labor and y units of capital will produce f(x, y) items where f(x, y) = k x n y 1 n, k and n are constants. If f(x, y) = 5x 0.7 y 0.3, how many units are produced using 8 units of labor and 10 units of capital? 7. A company produces and sells x pounds of chocolate at $p a pound and y pounds of fudge at $q a pound according to the demand functions: p = 20-3x + y q = 15 + x - 2y If the total cost C(x, y) = x +4y, find a) the Revenue function R( x, y) b) the Profit function P(x, y)

3 Chapter 7 page 3 FUNCTIONS OF SEVERAL VARIABLES A cigarette company promotes its product through magazine ads and by sponsoring sports events. The company estimates that if it spends x thousand dollars on magazines and y thousand dollars on sports events throughout the year, the company would expect to sell N thousand cartons of cigarettes per day, where N(x, y) = 25, x x 0.03y 2 +15y 1. If they are currently spending $10,000 on magazines and $20,000 on sports, what is their sales figure for N? (let x = 10 and y = 20) N( 10, 20) = 25, (10) (10) 0.03(20) 2 +15(20) = 25,768 so they expect to sell 25,768,000 cartons of cigarettes. 2. If they have an extra $1000 to put into either magazines or sports, where does this thousand do the most good? (in magazines or sports?) Spending the extra one thousand for magazines, the cigarette sales would be: N( 11, 20) = Spending the extra one thousand for sports, the cigarette sales would be: N( 10, 21) = This gives us two types of change in the number of cartons, N : * N(11,20) - N(10,20) = - =, which is the actual change in N when x increases by one, and y remains constant * N(10, 21) - N(10,20) = - =, which is the actual change in N when y increases by one, and x remains constant Where is the extra best spent? * NOTE: We can approximate and predict these changes using calculus >>>>>>>> >>>>>>>>>>>>>> turn this page over>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

4 Chapter 7 page 4 We can approximate and predict these changes using calculus: N(11,20) - N(10,20) = 45.8 can be approximated by: ΔN Δx PARTIAL DERIVATIVE of N with respect to x : notation: D x (N(x,y)) = N x (x,y) = N x = which is the Is the change in N as x increases by one and y remains constant They all mean : 1. The derivative of N with respect to x 2. The rate of change of N with respect to x 3. An estimate of the effect of a change of one unit in x on N 4. The marginal on N with respect to x. SO...N(11,20) - N(10,20) can be approximated by N x = -0.4x + 50 SO... N x (x, y) = 0.4x + 50 = N x (10, 20) = 0.4(10) + 50 = 46 thousand or 46,000 cartons N(10,21) - N(10,20) = can be approximated by: ΔN Δy PARTIAL DERIVATIVE of N with respect to x : notation: D y (N(x,y)) = N y (x, y) = N y = which is the Is the change in N as y increases by one and x remains constant They all mean : 1. The derivative of N with respect to y 2. The rate of change of N with respect to y 3. An estimate of the effect of a change of one unit in y on N 4. The marginal on N with respect to y. SO...N(10,21) - N(10,20) can be approximated by N y = -0.06y+15 SO... N y (x, y) = 0.06y +15 = N y (10, 20) = 0.06(20) +15 =13.8 thousand or13,800 cartons

5 Chapter 7 page PARTIAL DERIVATIVES part I 1. For z = 9 5x + 3y 2, find a) f x (x, y) = b) f x (3, 2) = c) f y (x,y) = d) f y (3, 2) = 2. For f(x,y) = x 3 2xy + y 2 + x 4y + 3, find a) f x = b) f x (3, 2) = c) f y = d) f y (2, 1) = 3. For C( x, y, ) = 2x 2 + 2xy + 3y 2 16x 18y + 54 find values of x and y for which both C x (x, y) = 0 and C y (x, y) = 0 4. If z = f(x,y) = 9 x 2 y 2, then a) f x (x, y) = b) f x (1,2) = c) f y (x,y) = d) f y (1,2) = 5. If f (x, y) = xe 2xy then find a) f x (x, y) = b) f y (x,y) = 6. f(x,y) = ln(5x 4y) a) f x (x,y) = b) f y (x,y) =

6 Chapter 7 page SEE PAGE 561: two paragraphs and illustration 3x 6y 7. f(x,y) = e a) f x (x,y) = b) f y (x,y) = 8. A company produces x standard items selling at $p each and y luxury items selling at $q each according to the demand functions p = 100-3x +y q = 80 + x - 2y with total cost function C(x,y) = x + 10y a) Find Revenue R(x,y) = b) Find R x (5,10) and R y (5,10) and interpret the results. c) Find Profit P(x,y) = d) Find P x (5,10) and P y (5,10) and interpret the results 9. A ski manufacturer has weekly production function S = f(x,y) = 250x 0.63 y 0.37 where S = number of pairs of skis produced using x units of labor and y units of capital. a) Find the marginal productivity of labor, S x = and the marginal productivity of capital, S y = b) Find f x (10, 20)= and f y (10, 20) = c) To increase productivity faster, should labor or capital be increased?

7 Chapter 7 page SECOND ORDER PARTIAL DERIVATIVES part II If z = f ( x, y ), then f xx = f xx (x, y) = 2 z x 2 = x ( z x ) f yy = f yy ( x, y) = 2 z y 2 = y ( z y ) Mixed Partial Derivatives: f xy = f xy (x, y) = 2 z y x = y ( z x ) First differentiate with respect to x, holding y constant. Then differentiate with respect to y, holding x constant f yx = f yx (x, y) = 2 z x y = y ( z y ) First differentiate with respect to y, holding x constant. Then differentiate with respect to x, holding y constant. 1. Find f xx, f xy, f yx,, and f yy for each function. a) f (x, y) = y 4 2xy 2 + 5x 2 +15x 8y b) f (x, y) = y x x y 2

8 Chapter 7 page SECOND PARTIAL DERIVATIVE TEST FOR EXTREMA OF f(x,y) 2nd Derivative Test for Local Extrema STEPS: 1. Find f x (x, y) and f y (x,y). 2. Set both = 0 and solve to find all critical points, (a,b). 3. Find all second partials: f xx, f yy, and f xy. 4. Plug in all critical values, one point at a time For each critical point, evaluate D = AC B 2 where A = f xx (a, b) B = f xy (a, b) C = f yy (a, b) I If D > 0 and A < 0, then f(a,b) is a RELATIVE MAXIMUM II If D > 0 and A > 0, then f(a,b) is a RELATIVE MINIMUM III If D < 0, the f(a,b) is a SADDLE POINT IV If D = 0, no information Most of the work in this extrema test is usually in the algebra needed to find the critical points. Some techniques occur frequently; a) If each partial contains only one variable: factor and solve. (see ex.1) b) If both partials are linear, use the addition (or elimination) method. (see ex.2) c) If one or both partials are nonlinear, use substitution from the easier partial into the harder one. (see ex.3) Examples: Classify all critical points of 1) f(x,y) = 2x 3 + y 2 9x 2 10y +12x 2 C.P. f xx = f yy = f xy = D conclusion

9 Chapter 7 page f(x,y) = x 2 + xy + y 2 4x 5y C.P. f xx = f yy = f xy = D conclusion 3. f(x,y) = 2x 3 6xy + 3y 2 + 6x 18y C.P. f xx = f yy = f xy = D conclusion

10 Chapter 7 page Applications of the Extrema Test 1. A company manufactures x ten-speed bicycles selling at $p and y three-speed bicycles selling at $q. The weekly demand and costs equations are p = 230-9x + y q = x - 4y C(x,y) = x + 30y a) What is the selling price for each bicycle when 10 ten-speed and 20 three-speed bicycles are sold? b) How many of each type of bicycle should be sold to maximize the weekly profit? c) What is the Maximum profit? MAYBE: A store sells two brands of color print film. The store pays $2 per roll of Brand A film and sells it for $p per roll. The store pays $3 per roll of Brand B film and sells it for $q per roll. The store sells x rolls of Brand A and y rolls of Brand B film each day according to the demand equations x = 75-40p + 25q y = p - 30q a) Determine the demands x and y when p = $4 and q = $5. b) How should the store price each brand to maximize the daily profit? c) What is the maximum profit?

11 Chapter 7 page MAXIMA AND MINIMA USING LAGRANGE MULTIPLIERS In the previous section we optimized functions of several variables. However, in many practical optimization problems, we must maximize or minimize a function in which the independent variable are subjected to certain further constraints. These extremum of a function f is called a constrained relative extremum of f. We will use a method called the Method of Lagrange Multipliers, named for Joseph Lagrange, a French mathematician. EXAMPLE: The Acrosonic Company manufactures a bookshelf loudspeaker system that may be bought fully assembled or in a kit. The total weekly profit (in dollars) that Acrosonic Company realized in producing and selling bookshelf loudspeaker systems is given by the profit function P(x,y) = 1 4 x2 3 8 y2 1 xy +120x + 100y where x denotes the number of fully assembled units and y denotes the number of kits produced and sold per week. Acrosonic s management decides the production of these loudspeaker systems should be restricted to a total of exactly 230 units per week. Under this condition, how many kits should be produced per week to maximize their weekly profit? ` The problem is equivalent to maximizing the function P(x,y) subject to the constraint x+y=230 To find this constrained relative maximum of P(x,y): we use the Method of Lagrange Multipliers: step1: Write the problem in the form Maximize (or minimize) z = f(x,y) P(x,y) = 1 4 x2 3 8 y2 1 xy +120x + 100y Subject to the constraint: g(x,y) =k g(x,y) = x + y = 230 The method of Lagrange multipliers uses the fact that any relative extremum of the function f(x,y)subject to the constraint f(x, y) = k must occur at a critical point (a, b) of the function F(x, y) = f (x, y) λ g(x, y) k [ ] where the new variable, λ, ( lambda, in honor of Lagrange) is called the Lagrange Multiplier step 2: Find the critical points for F by first finding its partial derivatives: F x = f x λg x F y = f y λg y F λ = (g k) step 3: Solve the equations F x = 0, F y = 0, and F λ = 0 simultaneously F x = f x λg x = 0 or f x = λg x F y = f y λg y = 0 or f y = λg y F λ = (g k) = 0 or g = k step 4: Evaluate f(a,b) at each critical point (a,b) of F

12 Chapter 7 page back to step 2: use the partial derivatives f x = f y = g x = g y = (step 3) to get the three Lagrange equations f x = λg x f y = λg y g = k Solving: The required constrained relative maximum of P occurs at the critical point (180, 50). Thus Acrosonic s profit is maximized by producing 180 assembled and 50 kit versions of their loudspeaker systems. The maximum weekly profit is P(180, 50) = $10, NOTE: If there had been more than one critical point, we would evaluate the Profit function for each critical point to determine the largest value. FOOTNOTE: λ = the number of additional units of the objective function for each additional additional unit of the constraint function or ΔP Δg

13 Chapter 7 page EXAMPLE 2: The total monthly profit of Robertson Controls Company in manufacturing and selling x hundred of its standard mechanical setback thermostats and y hundred of its deluxe electronic setback thermostats per month is given by the total profit function P(x,y) = 1 8 x2 1 2 y2 1 xy +13x + 40y where P is in hundreds of dollars. If the production of setback thermostats is to be restricted to a total of exactly 40 per month: a) how many of each model should the company manufacture in order to maximize its monthly profits? b) what is the maximum monthly profit?