Multivariable-calculus, logarithms I got the function f(x,y)=ln(1+x^2+y^2). There are three tasks to answer. a)Decide the function´s stationary points and classify them if possible. Here I got the answer to (0,0) is a local maximum point. b)Decide a Here I got the answer (0,0) c)Now limit the domain of definition to x^2+y^2<=1. Decide the function´s biggest and lowest value and the range. Please help me with task c). The answer should be: lowest value: f(0,0)=0. Biggest value: f(a,b)=ln(2) for all a^2+b^2=1. The range is 0<=z<=ln(2). How do I get there?

Landon Wolf

Landon Wolf

Answered question

2022-08-13

Multivariable-calculus, logarithms
I got the function f ( x , y ) = ln ( 1 + x 2 + y 2 ). There are three tasks to answer.
a)Decide the function´s stationary points and classify them if possible. Here I got the answer to ( 0 , 0 ) is a local maximum point.
b)Decide a Here I got the answer ( 0 , 0 )
c)Now limit the domain of definition to x 2 + y 2 1. Decide the function´s biggest and lowest value and the range.
Please help me with task c). The answer should be: lowest value: f ( 0 , 0 ) = 0. Biggest value: f ( a , b ) = ln ( 2 ) for all a 2 + b 2 = 1. The range is 0 z ln ( 2 ). How do I get there?

Answer & Explanation

Rowan Dyer

Rowan Dyer

Beginner2022-08-14Added 14 answers

f ( x , y ) = ln ( 1 + x 2 + y 2 )
Note that the function ln ( 1 + z ) is increasing with respect to z. Therefore ln ( 1 + x 2 + y 2 ) can be maximised or minimised be maximising or minimising x 2 + y 2 , respectively. x 2 + y 2 = 1 is the maximum value, and has infinitely many solutions, so the maximum of f ( x , y ) = ln ( 1 + 1 ) = ln 2 whenever x 2 + y 2 = 1. x 2 + y 2 = 0 is the minimum, and occurs only when x = 0 , y = 0. Therefore f ( x , y ) is minimised at f ( 0 , 0 ) = ln 1 = 0.

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