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if a function of one variable is continuous on an interval and has only one critical number, then a local maximum has to be an absolute maximum. But this is not true for functions of two variables. Show that the function f(x,y)=3xe^(y)-x^(3)-e^(3y) has exactly one critical point, and that f has a local maximum there that is not an absolute maximum. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible



Answer :

There is only one critical point for the provided function, and it has a local maximum there rather than an absolute maximum.

Explain the term Critical points?

  • The critical point is a location in the domain of a function where the derivative is either equal to zero or the function is not differentiable when dealing about functions of real variables.
  • Using the D(a,b), we may identify the important points of the a two-variable function and categorize them in Relative or Saddle Points.

The given function is;

f(x,y) = 3xe^(y) - x^(3) - e^(3y)

First partial derivatives:

fx = 3e^(y) - 3x^(2)

fy = 3xe^(y) - 3e^(3y)

Second partial derivatives:

f(xx) = -6x

f(yy) = 3xe^(y) - 9e^(3y)

f(yx) = 3e^(y)

Thus,

fx = 0 ⇒ 3e^(y) - 3x^(2) = 0

fy = 0 ⇒ 3xe^(y) - 3e^(3y) = 0

The possible critical point is: (1,0)

Second derivative test for:

Point: (1,0)

f(xx)(1,0) = -6

f(yy) (1,0) = -6

f(xy)(1,0) = 3

Now,

D(a,b) = f(xx)(a,b) . f(yy)(a,b) - [f(xy)(a,b)]²

Thus, (a,b)for (1,0);

D(1,0) = f(xx)(1,0) . f(yy)(1,0) - [f(xy)(1,0)]²

D(1,0) = -6  - 6  -9

D(1,0) = 27

Thus, the given function has only one critical point and has a local maximum there that is not an absolute maximum.

To know more about the Critical points, here

https://brainly.com/question/12476632

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