# Find the critical points on the given function f(x,y)=6xy+2,058/x+2,058/y?

find the critical points on the given function and classify each as a relative maximum, a relative minimum, or a saddle point.

f(x,y)=6xy+2,058/x+2,058/y

the critical point is ( ____,____) and it is a relative max or relative min?

pramodpandey | College Teacher | (Level 3) Valedictorian

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In continuation to above answer . To detrmine max /min ,we need to calculate `f_(x x) ,f_(y y) , and f_(xy)`

`f_(x x)=(2*2058)/x^3 ,{f_(x x)}_{(7,7)}=(2*2058)/7^3=12`

`f_(y y)=(2*2058)/y^3,{f_(y y)}_(7,7)=12`

`f_(xy)=6`

`{f_(x x)*f_(y y)-(f_(xy))^2}_(7,7)=12*12-6^2=108>0`

`{f_(x x)}_{(7,7)} >0`

`` Therefore (7, 7) will give local minima .

min f(x,y)=6*7*7+2*(2058)/7=882

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

`f(x,y)=6xy+2058/x+2058/y`           (i)

Find the partial derivative of (i) ,with respect to x and y respectively.

`f_x(x,y)=6y-2058/x^2`

`f_y(x,y)=6x-2058/y^2`

To obtain critical point of f(x,y) , set `f_x(x,y)=0 ,f_y(x,y)=0`

and solve resulting system of equations for x and y.

`6y-2058/x^2=0`

`6x-2058/y^2=0`

`x=343/y^2`

`y-343/(343/y^2)^2=0`

`y-(y^4)/343=0`

`343y-y^4=0`

`y(343-y^3)=0`

`y=0 , and y=7`

``But  y=0 then x can not be defined in second equation.

y=7 then x=7

so only critical point is (7,7)