Consider fx y x2 2xy y2 From the maen value theom there is

Consider f(x, y) = x^2 + 2xy +y^2, From the maen value theom, there is c = (minus t)0_2 +tz, 0 lessthan t lessthan 1 such that f(x) = f(z) minus f(0_2) Df z, where z = (x, y), 0_2 = (0,0)}. Find the value of t.

Solution

Df (p,q) = (2p+2q,2p+2q) at any point (p,q)

Now c= (tx,ty) t>0, 0<t<1

We require

x²+2xy+y² =Df(tx,ty) (x,y)

                 = 2t(x²+2xy+y2)

So t =1/2