Question

Examine the function f(x,y) = y2 + 4xy + 3x2 + x3 for extreme values

Answer 1

Step-by-step explanation:

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Naisha

Answer 2

Answer:

We have, f(x,y)=y2+4xy+3x2+x3

Step I:

fx=4y+6x+3x2

fy=2y+4x

fxx=6+6x

fxy=4

fyy=2

Step II:

We now solve fx=0, fy=0

simultaneously,

4y+6x+3x2=0

and 2y+4x=0

Putting 2y=−4x

in first equation

3x2−2x=0 ∴x(3x−2)=0

∴x=0

x=2/3

or When x=0, y=0

and when x=23, y=−43

∴(0,0), (23,−43)

are stationary points. **Step III**: (i) When x=0, y=0

r=fxx=6+6x=6+6(0)=6

s=fxy=4

t=fyy=2

\(\therefore rt-s^2 = 12 - 16 \lt 0\) We reject this pair (ii) When x=23, y=−43

r=fxx=10

s=fxy=4

t=fyy=2

\(\therefore rt-s^2 = 20 -16 \gt 0\) f(x,y)

is stationary at x=23, y=−43

But \(r=10\gt0\) Hence, f(x,y)

is minimum at x=2/3and y=-4/3