(b) Find the maximum and minimum values of the function f(x1,..., xn)subject to the constraint |||| = c where c E R>0 is a fixed positive real number.[Hint: The Lagrange multipliers algorithm applies in the same way to a function of nvariables as it does to functions of 2 or 3 variables. You may use, without proof, the factthat the set S = {T E R" : ||x|| = c} is closed and bounded and has no "edge points."]

Method of Lagrange multipliers: Maximize / minimize : f(x) Subject to the constraint : g(x)=0 Take a…

Answered: (b) Find the maximum and minimum values…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question 2: For which real values of a do the polynomials P1(t) = at? -t -; P2(t) = -t² + at – ; P3(t) = -t² -t + a form | a linearly dependent set in P2?
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Sections 2.4-2.6, 3.3, and 4.1- K Describe the end behavior of the polynomial function f. f(x) = 7x-2x³ Choose the correct answer below. O A. f(x) → ∞ as x → − ∞ and f(x) → ∞ as x → ∞ ○ B. f(x) – O C. OD. f(x) → - ∞ as x → - ∞ and f(x) - f(x) → → ∞ as x → Question 9 of 22 ∞ as x → − ∞ and f(x) → − ∞ as x → ∞ P x →∞ as X → ∞ Q Search − ∞ and f(x) → ∞ as x → ∞ DELL
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A1. Milk or meat production Y per day is best explained by the inputs: feed needed per day in kg, X₁, and (number of) milk cows or calves, X2. Based on fitting the data for milk or meat production the revenue is modelled by a Cobb-Douglas function R(X₁, X₂) = pX₁X₂ constrained by relation h(X₁, X2) = C₁ X₁ + C2X₂ = c3. For milk production, the constraint expresses a relation between the average sale price c3 of milk (per day), the average price c₁ of a kg of feed and the price c2 of a milk cow (cost per day and based on the annual cost of a milk cow). For meat production, the constraint expresses a relation between the average sale price c3 of meat (per day), the average price c₁ of a kg of feed and the price c₂ of a calf (cost per day and based on the annual cost of a calf). The parameters for the two cases have been established by linear regression of data to be the following: C1 £4.00, C₂ = C2 £1.36/d, c3 = £5.38; regarding the milk case: p = 445.69, a = 0.346, b = 0.542, c₁ = and,…
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The Intermediate Value Theorem can be used to approximate a root. The following is an example of binary search in computer science. Suppose you want to approximate /8. You know that it is between 2 and 3. If you consider the function f(x) = a? – 8, then note that f(2) 0. Therefore by the Intermediate Value Theorem, there is a value, 2 < c < 3 such that f(c) = 0. Next choose the midpoint of these two values, 2.5, which is guaranteed to be within 0.5 of the acutal root. f(2.5) will either be less than 0 or greater than 0. You can use the Intermediate Value Theorem again replacing 2.5 with the previous endpoint that has the same sign as 2.5. Continuing this process gives a sequence of approximations rn with T1 = 2.5. How many iterations must you do in order to be within 0.00390625 of the root?
can you amswer d,e?
The Intermediate Value Theorem can be used to approximate a root. The following is an example of binary search in computer science. Suppose you want to approximate /8. You know that it is between 2 and 3. If you consider the function f(x) = x? – 8, then note that f(2) 0. Therefore by the Intermediate Value Theorem, there is a value, 2 < c < 3 such that f(c) = 0. Next choose the midpoint of these two values, 2.5, which is guaranteed to be within 0.5 of the acutal root. f(2.5) will either be less than 0 or greater than 0. You can use the Intermediate Value Theorem again replacing 2.5 with the previous endpoint that has the same sign as 2.5. Continuing this process gives a sequence of approximations xn with x1 = 2.5. How many iterations must you do in order to be within 0.015625 of the root?

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