EXAMPLE 2 Find the extreme values of the function f(x, y) = 3x2 + 4y2 on the circle x2 + y2 = 1. SOLUTION We are asked for extreme values of f subject to the constraint g(x, y) = x2 + y2 = 1. Using Lagrange multipliers, we solve the equations ∇f = λ∇g and g(x, y) = 1, which can be written as fx = λgx fy = λgy g(x, y) = 1 or as (1) Correct: Your answer is correct. = 2xλ (2) Correct: Your answer is correct. = 2yλ (3) x2 + y2 = 1. From (1) we have x = 2 Incorrect: Your answer is incorrect. or λ = 3. If x = , then (3) gives y = ±1. If λ = 3, then y = from (2), so then (3) gives x = ±1. Therefore f has possible extreme values at the points (0, 1), (0, −1), (1, 0), and (−1, 0). Evaluating f at these four points, we find that

Question

23-03-2020
Mathematics

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EXAMPLE 2 Find the extreme values of the function f(x, y) = 3x2 + 4y2 on the circle x2 + y2 = 1. SOLUTION We are asked for extreme values of f subject to the constraint g(x, y) = x2 + y2 = 1. Using Lagrange multipliers, we solve the equations ∇f = λ∇g and g(x, y) = 1, which can be written as fx = λgx fy = λgy g(x, y) = 1 or as (1) Correct: Your answer is correct. = 2xλ (2) Correct: Your answer is correct. = 2yλ (3) x2 + y2 = 1. From (1) we have x = 2 Incorrect: Your answer is incorrect. or λ = 3. If x = , then (3) gives y = ±1. If λ = 3, then y = from (2), so then (3) gives x = ±1. Therefore f has possible extreme values at the points (0, 1), (0, −1), (1, 0), and (−1, 0). Evaluating f at these four points, we find that

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LammettHash LammettHash · Answer 1 · 2020-03-23T17:07:36+01:00

Without using Lagrange multipliers: Consider the same problem in polar coordinates, where we take

[tex]x=\cos\theta,y=\sin\theta[/tex]

to restrict [tex]f(x,y)[/tex] to the circle [tex]x^2+y^2=1[/tex], or [tex]r=1[/tex]. Then

[tex]f(x,y)=f(\cos\theta,\sin\theta)=3\cos^2\theta+4\sin^2\theta=3+\sin^2\theta[/tex]

which is a function of [tex]\theta[/tex] alone; denote this function by [tex]g(\theta)[/tex].

Find the critical points:

[tex]g'(\theta)=2\sin\theta\cos\theta=\sin(2\theta)=0\implies2\theta=n\pi\implies\theta=\dfrac{n\pi}2[/tex]

where [tex]n[/tex] is any integer. In one revolution of the circle, we get 4 critical points for [tex]n=0,1,2,3[/tex], which correspond to the points (1, 0), (0, 1), (-1, 0), and (0, 1) in Cartesian coordinates and to the extreme value of 3.

Take the second derivative:

[tex]g''(\theta)=2\cos(2\theta)[/tex]

Then [tex]g''(0)[/tex] and [tex]g''(\pi)[/tex] are both positive, while [tex]g''\left(\frac\pi2\right)[/tex] and [tex]g''\left(\frac{3\pi}2\right)[/tex] are both negative, which indicates (1, 0) and (-1, 0) are minima, and (0, 1) and (0, 1) are maxima.