Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point.

critical point of the given function f(x,y) = x²+y²+2xy is from line y = -x is the critical point of the function f(x0,y0) = 0

and it local minimum.

Step-by-step explanation:

Let the given function be;

f(x,y) = x²+y²+2xy

From above function, we can locate relative minima, maxima and the saddle point

f(x,y) = x²+y²+2xy = (x+y)²

df/dx = 2x+2y = 0 (1)

df/dy =2y+2x = 0 (2)

From eqn 1 and 2 above,

The arbitrary point (x0,y0) from line y = -x is the critical point of the function f(x0,y0) = 0

Then, from f(x,y) >= 0 for arbitrary (x,y) € R^n, the arbitrary point from the line x = -y is local minima of the function f.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.