ok, hello!
Determine all local minimum, maximum and saddle points for .
This is not a question of "find the max and min of y as a function of x", it is "find the max and min of f(x,y)". That is, think of z= f(x,y) and find critical points of z.
The critical points of a function of two variables are where which is the same as saying [tex]\frac{\partial f}{\partial x}= 0[tex] and .
The "second derivative test" for a function of two variables is more complicated than for a function of one variable. First calculate
.
If D> 0, then the point is a maximum if is positive and a minimum if it is negative. If D< 0, then the point is a saddle point.
Ah yes! Thanks HallsofIvy,
A critical point occurs when both partial derivatives are zero.
Hence, we solve the resulting pair of equations for x and y.
The critical points are (0,0) and (-5,5).
To determine the nature of the critical points, the second partial derivatives are examined.
Also required is
D>0 and relative minimum point
D>0 and relative maximum point
D<0 saddle point
At (0,0) D<0 therefore this is a saddle point.
At (-5,5) D is 900-225 >0 and so this is a relative maximum.
The 2nd derivative of y is also utilised,
the results for "D", the "discriminant" are determined by the equation for D,
which is regarded as a theorem.
The equation for D does involve the product of both 2nd derivatives.
The saddle points have a "local maximum for x" and a "local minimum for y" colliding at a point or vice versa, which is a way of putting it.
Hence no maximum for both x and y simultaneously (or maximum for both) to identify the point as either a maximum or minimum.
D<0 for a saddle point.
If D>0, then a critical point will be a maximum or minimum.
The reason is based on the formula for D.
Since the RHS of the above expression is always positive,
then the only way D can be >0 is for both 2nd derivatives to be simultaneously positive or simultaneously negative.
Hence, if D>0, there is no need to double-check both 2nd derivatives.