In Calculus we can find the critical points for a given equation. Critical points are those points for which the function has zero value, when solved around the given variables, used in the equation. We do this by calculating the differentiation for those points and then evaluating those equations with zero.
If we have a Point say K0 = (b1, b2, b3....) is a maximum point or a minimum point of the equation y= f(x1, x2, x3....xn ) and we can certain all the values apart from x1 which is in relation with a1.
In such case we can say that df/dx1 = 0. While fx1 , fx2, fx3 are all equal to zero at point P.
Let us understand this with the help of an example:
f(z) = z5 +x4 - 5z – 32x
Now, we have to find:
df / dz = 5z4 - 5
df / dx = 4x3 -32 = 0
Thus solving for the critical points, we get ( 1, 2 ) and ( -1 , 2 )
Now, after the calculation of these points let us see, where they lie the maximum and the minimum value.
For point (1, 2) the equation , we will notice that for z = 1the equation is minimum and for point x= 2 it is minimum for the second equation. Therefore the point (1,2) is minimum for f(z,x) = z5 +x4 - 5z – 32x. While the point z = -1 is a maximum for the first equation. The critical point of ( -1, 2) in neither maximum nor a minimum point for the surface. It is a saddle point.
In a Similar way, we can also solve the function for the new equation where we have variables in the mixed form that is: f(p,q) = p2 +2pq +2q2 + +2p -2p. In this as well we have critical points : fp = 0 and fq =0. Here also we will simplify this problem in the similar way, this gaining the critical points to be (-3, 2).
Now, we will calculate the second derivative to see if the critical points are satisfied or not.
Here, fpp . Fqq – (fpq)2 = 8-4 = 4 >0.
The second derivative shows that at the point ( -3, 2) we have a local minimum. Thus we have learn finding critical points multivariable calculus