Math 2 Spring 2 Final Eam Practice Problems (Set 2) 1. Find the etreme values of f(, ) = 2 2 + 3 2 4 on the region {(, ) 2 + 2 16}. 2. Allocation of Funds: A new editor has been allotted $6, to spend on the development and promotion of a new book. It is estimated that i thousand dollars is spent on development and thousand on promotion, approimatel f(, ) = 2 3/2 copies of the book will be sold. How much mone should the editor allocate to development and how much to promotion in order to maimie sales? 3. Marginal Analsis: Suppose the editor in the earlier problem is allotted $61, instead of $6, to spend on the development and promotion of the new book. Estimate how the additional $1, will affect the maimum sales level.
4. Match each of the following functions with its graph and with its contour map. (a) f(, ) = e cos (b) f(, ) = 2 3 (c) f(, ) = 3 3 (d) f(, ) = sin 2 + 2 Graphs A B C D Contour Maps 1 I II 7. 2. -2. -7. -1-1 1-7. -2. 2. 7. 1 III 6 IV 4 2-2 -4-1 -1 1-6 -2 2 4 6
Answers 1. Here s a similar problem from first-semester calculus: Find the etreme values of the function f() = 1 (2 4 12) on the interval [ 4, 4]. 6 4 2-4 -2 2 4-2 The etreme values might occur at the endpoints of the interval or in the interior of the interval. If an etreme value occurs in the interior of the interval, then it must occur at a critical point. For the above function, the absolute maimum occurs at the endpoint = 4 and the absolute minimum occurs at the critical point = 2. To find these etreme, ou would have to first find the critical points and then evaluate the function at all critical points and endpoints. We will use a similar strateg for the two-variable problem at hand. The etreme values of the function f(, ) = 2 2 + 3 2 4 on the region {(, ) 2 + 2 16} will occur either on the interior of the region or on the boundar of the region. That is, the will either occur on the interior {(, ) 2 + 2 < 16} or on the boundar {(, ) 2 + 2 = 16}. Step One: Determine the critical points of the function on the interior of the region. Note that = 4 4 and f = 6. Setting both of these equal to ero, we find that the onl critical point is = 1, =. The point (1, ) is in the interior {(, ) 2 + 2 < 16}, so it is a candidate for the etreme values of this function on the region {(, ) 2 + 2 16}. Step Two: If the etreme values do not occur at the critical point of the function, then the must occur at points on the boundar of the region, {(, ) 2 + 2 = 16}. Since there are infinitel man such points, we don t want to test them all! We ll use the technique of Lagrange multipliers to determine which of the boundar points are etreme points on the boundar. This means that we want to identif the etreme points of the function f(, ) = 2 2 + 3 2 4 subject to the constraint 2 + 2 = 16. This gives us our function g: g(, ) = 2 + 2 16. The etreme points will occur at (, ) pairs satisfing the following two equations. Simplifing the first equation, we get which implies that either = or = 2. and g(, ) = 4 4 2 = 6 2 (4 4)(2) = (6)(2) (4 4)(2) (6)(2) = (4)(( 1)(2) (3)()) = (4)( 2) =
Case One: If =, then we can substitute that in the g equation to get = 2 + 2 16 = 2 + 2 16 = 2 16 which implies that = ±4. This gives us two candidates for etreme values: ( 4, ) and (4, ). Case Two: I = 2, then we can substitute that in the g equation to get = 2 + 2 16 = ( 2) 2 + 2 16 = 2 12 which implies that = ± 12. This gives us two more candidates for etreme values: ( 2, 12) and ( 2, 12). Step Three: We now have five candidates for etreme values. We ll substitute each of them into f to see which give the greatest outputs and which give the least outputs. f(1, ) = 7 f( 4, ) = 43 f(4, ) = 11 f( 2, 12) = 47 f( 2, 12) = 47 Thus, the function obtains its maimum value of 47 at the points ( 2, 12) and ( 2, 12) and it obtains is minimum value of 7 at the point (1, ). 2. Note that we want to maimie the function f(, ) = 2 3/2 subject to the constraint + = 6. This gives us our g function: g(, ) = + 6. The etreme points will occur at (, ) pairs satisfing the following two equations. Simplifing the first equation, we get which implies that either = or = 2 3. and g(, ) = 3 1/2 = 23/2 1 1 3 1/2 = 2 3/2 3 1/2 2 3/2 = 1 1/2 (3 2) = Case One: I =, then we can substitute that in the g equation to get = + 6 = + 6 which implies that = 6. This gives us one candidate for the maimum value: (, 6). Case Two: If = 2 3, then we can substitute that in the g equation to get = + 6 = + 2 3 6 = 3 6 which implies that = 36. Since = 2 3, we get that = 2 3 36 = 24. This gives us another candidate for the maimum value: (36, 24).
We ll substitute each of these into f to see which give the greatest output. f(, 6) = f(36, 24) = 13, 68 Thus, we can maimie sales b spending $36, on development and $24, on promotion, ielding sales of 13,68 books. 3. Recall the following two items from the lecture notes on Lagrange multipliers. Lagrange Multipliers: Given a point (, ) that satisfies the two equations let We call λ a Lagrange multiplier. and g(, ) =, λ =. Theorem: Suppose M is the maimum (or minimum) value of f(, ) subject to the constraint g(, ) = k. The Lagrange multiplier λ is the rate of change of M with respect to k. That is, λ = dm dk. Hence λ approimates the change in M resulting from a 1-unit increase in k. In this problem M is the maimum sales corresponding to a budget of k thousand dollars. Note that the instantaneous rate of change in M with respect to k is approimatel equal to the average rate of change in M with respect to k: dm dk M k, assuming k is relativel small. This gives us that or λ = dm dk M k M λ k. This tells us that if we change the budget b k thousand dollars, then the change in maimum sales is approimatel λ k. In this eample, k = 1 and λ = = 31/2 =36,=24 1 = 3(36) 1/2 (24) = 432. =36,=24 Thus, b increasing the budget b one thousand dollars, we can epect to sell 4,32 more books. 4. (a) D, IV (b) B, I (c) A, III (d) C, II