Unveiling Critical Points And Extremes: A Deep Dive Into Multivariable Calculus

Hey math enthusiasts! Today, we're going to dive headfirst into the fascinating world of multivariable calculus. We'll be tackling a classic problem that involves finding critical points, classifying them as minima, maxima, or saddles, and then proving the existence (or non-existence) of global extremes. Buckle up, because we're about to embark on a mathematical adventure!

(a) Finding and Classifying Critical Points

Alright, guys, let's get down to business. We're given the function f(x, y) = 2x³ - y² + 6x² + 4y + 11. Our first mission? Finding all the critical points of this function. Remember, critical points are the points where the gradient of the function is either zero or undefined. In our case, since f(x, y) is a polynomial, it's defined everywhere, so we just need to find where the gradient is zero. The gradient is a vector of partial derivatives, so we need to calculate those first.

Step 1: Calculate the Partial Derivatives

Let's get those partial derivatives rolling. We'll start with the partial derivative of f with respect to x, denoted as fₓ:

fₓ(x, y) = ∂f/∂x = 6x² + 12x

Next up, the partial derivative of f with respect to y, denoted as fᵧ:

fᵧ(x, y) = ∂f/∂y = -2y + 4

Step 2: Set the Partial Derivatives to Zero

To find the critical points, we need to solve the system of equations where both partial derivatives are equal to zero. So, we have:

• 6x² + 12x = 0*
• -2y + 4 = 0*

Step 3: Solve the System of Equations

Let's solve these equations. The first equation, 6x² + 12x = 0, can be factored as 6x(x + 2) = 0. This gives us two possible values for x: x = 0 and x = -2.

The second equation, -2y + 4 = 0, is easy to solve. It gives us y = 2.

Therefore, we have two critical points: (0, 2) and (-2, 2). Awesome, we're halfway there! Now we need to figure out what kind of points these are.

Step 4: Classify the Critical Points (Second Derivative Test)

To classify these critical points, we'll use the second derivative test. This test involves calculating the determinant of the Hessian matrix, which is a matrix of second partial derivatives. Let's find those second partial derivatives:

• fₓₓ(x, y) = ∂²f/∂x² = 12x + 12
• fᵧᵧ(x, y) = ∂²f/∂y² = -2
• fₓᵧ(x, y) = ∂²f/∂x∂y = 0 (and fᵧₓ(x, y) = 0, they're equal for continuous functions, which ours is)

The Hessian matrix, H, is:

H = [[fₓₓ, fₓᵧ], [fᵧₓ, fᵧᵧ]] = [[12x + 12, 0], [0, -2]]

The determinant of the Hessian matrix, D, is:

D = det(H) = (12x + 12)(-2) - (0)(0) = -24x - 24

Now, we'll evaluate D at each critical point.

• At (0, 2): D = -24(0) - 24 = -24. Since D < 0, this is a saddle point.
• At (-2, 2): D = -24(-2) - 24 = 24. Also, we need to check fₓₓ(-2, 2) = 12(-2) + 12 = -12. Since D > 0 and fₓₓ < 0, this is a local maximum.

So, we've classified our critical points! (0, 2) is a saddle point, and (-2, 2) is a local maximum. Nice work, team!

(b) Proving the Absence of a Global Maximum or Minimum

Alright, let's tackle the second part of our challenge: proving that f(x, y) has no global maximum or minimum on R². This might seem a bit tricky at first, but with a little intuition, we can make quick work of it. We are going to make use of limits to show that there's no global max or min.

The Intuitive Approach: Exploring the Behavior of the Function

To understand why there's no global max or min, let's think about the behavior of the function f(x, y) = 2x³ - y² + 6x² + 4y + 11. Notice that the term 2x³ dominates the function's behavior as x becomes very large (positive or negative). The x³ term will be bigger than the other terms when x is big enough. If x goes to positive infinity, f(x, y) will also go to positive infinity. Likewise, if x goes to negative infinity, f(x, y) will go to negative infinity.

The Limit Proof

Let's formalize this with limits. First, to show there's no global maximum, we'll take the limit as x goes to positive infinity:

lim (x→+∞) f(x, y) = lim (x→+∞) (2x³ - y² + 6x² + 4y + 11). The y² and 4y don't change the behavior here, since x³ is so much bigger than them.

Since the 2x³ term goes to positive infinity as x goes to positive infinity, the entire function goes to positive infinity. Therefore, f(x, y) has no global maximum.

Next, to show there's no global minimum, we'll take the limit as x goes to negative infinity:

lim (x→-∞) f(x, y) = lim (x→-∞) (2x³ - y² + 6x² + 4y + 11).

As x goes to negative infinity, 2x³ goes to negative infinity. Thus, the entire function goes to negative infinity. Therefore, f(x, y) has no global minimum.

Conclusion

And there you have it! Because the function approaches positive and negative infinity, we can conclude that f(x, y) has no global maximum or minimum on R². The existence of a saddle point also gives further proof that there is no global min or max, but is not needed in the solution of the problem.

Conclusion

So, there you have it! We've successfully found and classified the critical points of f(x, y) and shown that it has no global maximum or minimum. This problem showcases some key concepts in multivariable calculus, and I hope this walkthrough has helped solidify your understanding. Keep practicing, and you'll become a multivariable calculus master in no time! Keep exploring, keep questioning, and keep having fun with math, everyone! Until next time!