Question

Question:
Examine the function f(x, y) = 2x2 + y4 – 4xy for extremum.

Answer 1

Step-by-step explanation:

f(x)=x4+y4−2(x2+y2)+4xy

change of variables (polar coordinates):

x=rcos(t)y=rsin(t)

f(t)=r4(cos4(t)+sin4(t))−2r2(cos2(t)+sin2(t))+4r2cos(t)sin(t)

f(t)=r4(cos2(t)+sin2(t))2−2r4cos2(t)sin2(t)−2r2(cos2(t)+sin2(t))+2r2sin(2t)

f(t)=r4−r4sin2(2t)2−2r2+2r2sin(2t)

⇒f(t)=(2r4−4r2−(r4sin2(2t)−4r2sin(2t))2

⇒f(t)=2(r2−1)2+2−(r2sin(2t)−2)22

It is clear that f has no maximal value

The minimal value occurs when:

(r2sin(2t)−2)2 is maximal and equal to (−r2−2)2=(r2+2)2

that is when: sin(2t)=−1

and f(t)=2(r2−1)2+2−(r2+2)22=r4−8r22

this quantity is the minimal value on the circle with radius r

Then the absolute minimum is:

minr4−8r22( remember that: r≥0)

this minimu occures when: 4r3−16r=0 that is r=2

sin(2t)=−1⇒2t=−π2⇒t=−π4

x=rcos(t)=2cos(−π4)=2–√

y=−2–√The minimal value occurs when:

(r2sin(2t)−2)2 is maximal and equal to (−r2−2)2=(r2+2)2

that is when: sin(2t)=−1

and f(t)=2(r2−1)2+2−(r2+2)22=r4−8r22

this quantity is the minimal value on the circle with radius r

Then the absolute minimum is:

minr4−8r22( remember that: r≥0)

this minimu occures when: 4r3−16r=0 that is r=2

sin(2t)=−1⇒2t=−π2⇒t=−π4

x=rcos(t)=2cos(−π4)=2–√

y=−2–√

the minimum value of f is:

r4−8r22=16−8×42=−8