Find the relative extreme value of the function.  f(x,y)=2xy-2x^2-3y^2+4x-12y+5  see given relative extrema.Given: relative extrema test: D=fxxfyy - fxy  ^2 D>0 and fxx> 0 --> relative...

Find the relative extreme value of the function. 

f(x,y)=2xy-2x^2-3y^2+4x-12y+5 

see given relative extrema.

Given: relative extrema test:

D=fxxfyy - fxy  ^2

D>0 and fxx> 0 --> relative min.

D>0 and fxx <0 --> relative max.

D<0 --> saddle point

*for fxx;fyy;fxy in all forms above xx,yy, and xy are all small.

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lfryerda | High School Teacher | (Level 2) Educator

Posted on

The relative extrema exist where the first partial derivatives vanish.  To determine the nature of the relative extrema, we calculate the value of D.

So, we need to calculate the partial derivatives to find D.

since `f=2xy2x^2-3y^2+4x-12y+5`

then `f_x=2y-4x+4`

and `f_{x x}=-4`

Also, `f_y=2x-6y-12`

and `f_{y y}=-6`

Finally, `f_{x y}=2` .

This means that `D=f_{x x} f_{y y}- f_{x y} ^2 = (-4)(-6)-4=20>0` , which will lead to a realtive min for the extrema since `f_{x x} = -4 <0`. 

The relative min are where the first partial derivatives `f_x=2y-4x+4` and `f_y=2x-6y-12` vanish.

This means that the relative mins are found on the lines `y=2x-2` (from `f_x=0`) and `y=1/3x-4` (from `f_y=0`).

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