Answer
The local extrema is 4.
Work Step by Step
Consider f(x,y) = x^{2}*y
g(x,y) = x+y-3=0
Hence, f(x,y) = (x^{2})*y + \lambda * (x + y - 3)
\frac{df}{dx} = 2*x*y - \lambda
\frac{df}{dy} = x^{2} - \lambda
\frac{df}{d\lambda} = -x-y+3
Now,
\frac{df}{dx} = 0
\frac{df}{dy} = 0
\frac{df}{d\lambda} = 0
Hence, by solving the equation we find,
x=2
y=1
Thus, the extremum point is (2, 1).
So, f (2,1) = (2^{2}) * 1=4
Hence, the local extrema is 4.
