Answer:
To investigate the maxima and minima of these functions, we can start by finding the partial derivatives with respect to x and y, setting them equal to zero, and solving for x and y.
(i) Starting with 21x12x²-2y² + x² + xy², the partial derivatives are:
∂f/∂x = 42x^3 + 2xy ∂f/∂y = -4y + 2xy²
Setting these equal to zero and solving for x and y, we get:
42x^3 + 2xy = 0 (equation 1) -4y + 2xy² = 0 (equation 2)
From equation 1, we can solve for y in terms of x:
y = -21x^2/2
Substituting this into equation 2, we get:
-4(-21x^2/2) + 2x(-21x^2/2)^2 = 0
Simplifying and solving for x, we get:
x = 0 or √(2/63)
Plugging these values into the original function and evaluating, we get:
f(x=0, y=0) = 0 f(x=√(2/63), y=-1/6√2) ≈ -0.027
So the global minimum of this function occurs at (x=√(2/63), y=-1/6√2), and the value of the function at that point is approximately -0.027. There are no local maxima or minima.
(ii) For 2(x-y)²-x² - y², the partial derivatives are:
∂f/∂x = -2x + 4(x-y) ∂f/∂y = -2y + 4(y-x)
Setting these equal to zero and solving for x and y, we get:
x = y x = 2y
These equations are inconsistent, so there are no critical points. The function has no local maxima or minima.
(iii) For x² + 3xy + y² + x² + y², the partial derivatives are:
∂f/∂x = 2x + 3y ∂f/∂y = 3x + 2y
Setting these equal to zero and solving
Explanation: