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Application of Partial Derivative – Two variable Maxima and Minima

  • Last Updated : 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,

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  • 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 :



  1. 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.]
  2. The Obtained result will be considered as stationary/turning points for the curve.
  3. Create 3 new variables r,t and s.
  4. Find the values of r,t and s using r=fxx, t=fyy, s=fxy
  5. If (rt-s2)|(stationary pts)>0  (Maxima/Minima) exists
  6. If (rt-s2)|(stationary pts) <0 (No Maxima/Minima)/(Saddle point)
  7. If r=fxx>0 (Minima) 
  8. 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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