Question

Find local extremes of f(s,t)=s^4 +t^4 −4st+1.

A) Local maximum at (0, 1) and local minimum at (1, 0).
B) Local maximum at (1, 0) and local minimum at (0, 1).
C) Local maximum at (1, 1) and local minimum at (0, 0).
D) Local maximum at (0, 0) and local minimum at (1, 1).

Answer 1

To find the local extremes of f(s,t)=s^4 + t^4 - 4st + 1, we need to find the critical points and perform the second partial derivative test. The local maximum is at (1,0) and the local minimum is at (0,1).

Step-by-step explanation:

To find the local extremes of f(s,t)=s^4 + t^4 - 4st + 1, we need to find the critical points where the partial derivatives equal zero. Taking the partial derivative with respect to s, we get 4s^3 - 4t = 0.

Taking the partial derivative with respect to t, we get 4t^3 - 4s = 0.

Solving these two equations simultaneously, we find that the critical points are (0,1) and (1,0). To determine if these points are local maxima or minima, we can use the second partial derivative test.

Taking the second partial derivatives, we get 12s^2 and 12t^2, which are both positive.

Therefore, the point (0,1) is a local minimum, and the point (1,0) is a local maximum. So the correct answer is option A) Local maximum at (0, 1) and local minimum at (1, 0).