Letf=x2 +y2(1+x)3 (a) Show that f has only one critical point and that it is a local minimum at that critical point. (b) Show that f has no minimum value.

Question

Let f = x^2 +y^2 (1+x)^3

(a) Show that f has only one critical point and that it is a local minimum at that critical point.

(b) Show that f has no minimum value.

Answer 1

f(x. y) = x² + y²(1 + x)³.

(a) ∂f(x,y)/∂x = 2x + 3y²(1 + x)² = 0;

∂f(x,y)/∂y = 2y(1 + x)³ = 0;

If y - 0 then x = 0. So, (0, 0) is critical point.

If (1 + x)³ = 0, then x = - 1 and first equation is contradiction.

∂²f/∂x² = 2 + 6y²(1 + x); ∂²f/∂x² at (0, 0) equal 2

∂²f/∂y² = 2(1 + x)³; ∂²f/∂y² at (0, 0) equal 2

∂²f/∂x∂y = 6y(0.1 + x)²; ∂²f/∂x∂y at (0, 0) = 0.

So, D(0, 0) -(∂²f/∂x²·∂²f/∂y² - (∂²f/∂x∂y)²)(0, 0) = 2·2 - 0² = 4 > 0 and because at (0, 0) f(x, y) has minimum;

min f = f(0, 0) = 0;

(b) This function has no absolute minimum because if for example y = x, then f(x, x) = x² + x²)(1 + x)³=

x⁵ + 3x⁴ + 3x³ + 2x² → - ∞ as x→ - ∞