# Application of Partial Derivative – Two variable Maxima and Minima

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 :**

- Find the values of x and y using
**f**and_{xx}=0**f**[_{yy}=0**NOTE**: f_{xx }and f_{yy}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=f**_{xx,}t=f_{yy}, s=f_{xy} **If**(**rt-s**^{2})|_{(stationary pts)}>0**(Maxima/Minima) exists****If (rt-s**^{2})|_{(stationary pts)}<0 (No Maxima/Minima)/(Saddle point)- If
**r=f**_{xx}>0 (Minima) - If
**r=f**_{xx}<0 (Maxima)

**Example-1 :**

The function f(x,y)=x^{2}y−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=∂^{2}f/∂x^{2}=2y s=∂^{2}f/∂x∂y=2x−3 t=∂^{2}f/∂y2=0

Since, rt−s^{2}≤0, (if rt-s^{2}< 0 then we have no maxima or minima, if = 0 then we can’t say anything).

Maxima will exist when rt−s^{2}>0 and r<0.

Minima will exist when rt−s^{2}>0 and r>0.

**As rt−s ^{2} is never greater than 0 so we have no local extremum.**

**Example-2 :**

Find the local minima of the function f(x , y) = 2x^{2} + 2xy + 2y^{2} – 6x

f_{x}(x,y) = 4x + 2y - 6=0 (1) f_{y}(x,y) = 2x + 4y=0 (2)

On solving (1) and (2) we get,

x=2,y=-1r=∂^{2}f/∂x^{2}=4 s=∂^{2}f/∂x∂y=2 t=∂^{2}f/∂y^{2}=4 rt−s^{2}=12

**As rt−s ^{2}>0 and r>0. Thus, (2,-1) is the point of local minima.**

**Example-3 :**

Find the maxima/minima of f(x , y) = x^{2}+y^{2} + 6x +12

f_{x}(x,y) = 2x+6=0 (1) f_{y}(x,y) = 2y=0 (2)

On solving (1) and (2) we get,

x=-3,y=0r=∂^{2}f/∂x^{2}=2 s=∂^{2}f/∂x∂y=0 t=∂^{2}f/∂y^{2}=2

**As rt−s ^{2}>0 and r>0. Thus, (-3,0) is the point of local minima.**