We have been given a function f(x,y) = 7x^2 - 6y^2 - 28x - 36y - 5 and the task is to determine its relative extrema and saddle points.
Step 1: We need to find the critical points for the function, which are the points where the first derivative of the function is zero. Hence, we need to find the first partial derivatives of the function with respect to 'x' and 'y.'
The first partial derivative of f with respect to x is:
fx = 14x- 28
Setting fx = 0 gives x =2.
Similarly, the first partial derivative of f with respect to y is:
fy = -12y - 36
Setting fy = 0 gives y = -3.
Therefore, the critical point of this system is at (2, -3).
Step 2: To classify this critical point (x0, y0) as a local minimum, local maximum or saddle point, we use the second derivative test. This requires the second partial derivatives of f.
The second partial derivatives of f are:
f x x = 14 (second derivative with respect to x)
f y y = -12 (second derivative with respect to y)
f x y = 0 (mixed derivative)
The discriminant D of the function is defined as D = fx x*fyy - (fxy)^2 = 14*(-12) - (0)^2 = -168.
Step 3: Now, we can use the discriminant D to classify the critical point.
If D > 0 and (f x x > 0), then f has a local minimum at (x0, y0)
If D > 0 and (f x x < 0), then f has a local maximum at (x0, y0)
If D < 0, then f has a saddle point at (x0, y0)
For the point (2,-3), since D = -168 < 0, we conclude that (2,-3) is a saddle point of f.
Now you should continue these steps for the other reference points: (−2,−3,133), (2,3,−195) and repeat the procedure. Note that if you substitute these points into the original equation and the discriminant remains less than zero, they will also be saddle points. If the result is more than zero, they could potentially be relative minimums or maximums, depending on the value of (f x x). Please follow steps 1-3 to classify these points.