Application of Partial Derivative – Two variable Maxima and Minima - GeeksforGeeks

22 Oct, 2021
Partial derivatives can be used to find the maximum and minimum value (if they exist) of a two-variable function. We try to locate a stationary point that has zero slope and then trace maximum and minimum values near it. The practical application of maxima/minima is to maximize profit for a given curve or minimizing losses.
Let f(x,y) be a real-valued function and let (pt,pt’) be the interior points in the domain of f(x,y) then,
pt, pt’ is called a point of local maxima if there is an h > 0 such that f(pt,pt’) ≥f(x,y), for all x in (pt – h, pt’ + h), x≠a The value f(pt,pt’) is called the local maximum value of f(x,y).
pt, pt’ is called a point of local minima if there is an h < 0 such that f(pt,pt’) ≥f(x,y), for all x in (pt – h, pt’ + h), x≠a The value f(pt,pt’) is called the local minimum value of f(x,y).
Algorithm to find maxima and minima of two-variable functions :
Find the values of x and y using fxx=0 and fyy=0  [NOTE: fxx and fyy are the partial double derivatives of the function with respect to x and y respectively.]The Obtained result will be considered as stationary/turning points for the curve.Create 3 new variables r,t and s.Find the values of r,t and s using r=fxx, t=fyy, s=fxyIf (rt-s2)|(stationary pts)>0  (Maxima/Minima) existsIf (rt-s2)|(stationary pts) <0 (No Maxima/Minima)/(Saddle point)If r=fxx>0 (Minima) If r=fxx<0 (Maxima)
Find the values of x and y using fxx=0 and fyy=0  [NOTE: fxx and fyy are the partial double derivatives of the function with respect to x and y respectively.]
The Obtained result will be considered as stationary/turning points for the curve.
Create 3 new variables r,t and s.
Find the values of r,t and s using r=fxx, t=fyy, s=fxy
If (rt-s2)|(stationary pts)>0  (Maxima/Minima) exists
If (rt-s2)|(stationary pts) <0 (No Maxima/Minima)/(Saddle point)
If r=fxx>0 (Minima) 
If r=fxx<0 (Maxima)
Example-1 :
The function f(x,y)=x2y−3xy+2y+x has                
(a) No local extremum
(b) One local minimum but no local maximum
(c) One local maximum but no local minimum
(d) One local minimum and one local maximum
Explanation :
Answer: A
r=∂2f/∂x2=2y
s=∂2f/∂x∂y=2x−3
t=∂2f/∂y2=0
Since, rt−s2≤0, (if rt-s2< 0 then we have no maxima or minima, if = 0 then we can’t say anything).
Maxima will exist when rt−s2>0 and r<0.
Minima will exist when rt−s2>0 and r>0.
As rt−s2 is never greater than 0 so we have no local extremum.
Example-2 :
Find the local minima of the function f(x , y) = 2x2 + 2xy + 2y2 – 6x
fx(x,y) = 4x + 2y - 6=0    (1)
fy(x,y) = 2x + 4y=0        (2)
On solving (1) and (2) we get,
x=2,y=-1
r=∂2f/∂x2=4
s=∂2f/∂x∂y=2
t=∂2f/∂y2=4
rt−s2=12
As rt−s2>0 and r>0. Thus, (2,-1) is the point of local minima.
Example-3 :
Find the maxima/minima of  f(x , y) = x2+y2 + 6x +12
fx(x,y) = 2x+6=0     (1)
fy(x,y) = 2y=0       (2)
On solving (1) and (2) we get,
x=-3,y=0
r=∂2f/∂x2=2
s=∂2f/∂x∂y=0
t=∂2f/∂y2=2
As rt−s2>0 and r>0. Thus, (-3,0) is the point of local minima.
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Engineering Mathematics
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