91Ó°ÊÓ

In multivariable calculus, finding critical points is essential for determining where a function has potential extrema (maxima or minima). A critical point occurs when the partial derivatives of a function are zero, indicating a horizontal tangent plane. In simpler terms, the slope in all directions is zero.
To locate the critical points of a function like \( f(x, y) = 4y + 6x - x^2 - y^2 \), we start by computing its partial derivatives:

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 6 - 2x \).
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 4 - 2y \).

By setting these partial derivatives to zero, \( 6 - 2x = 0 \) and \( 4 - 2y = 0 \), we solve for \( x \) and \( y \). This gives us the critical point \((3, 2)\). Here, both slopes (or changes) in the \(x\) and \(y\) directions are zero, making it a candidate for a relative extremum.

Determining the nature of critical points further involves the second derivative test. This test reveals whether the critical point is a minimum, maximum, or a saddle point. The test uses the second partial derivatives to form a Hessian matrix and its determinant, which provides crucial insight.
For the function given, we compute:

• \( \frac{\partial^2 f}{\partial x^2} = -2 \)
• \( \frac{\partial^2 f}{\partial y^2} = -2 \)
• The mixed second partial derivative \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

The determinant of the Hessian matrix \( D \) is calculated as:\[ D = \left( \frac{\partial^2 f}{\partial x^2} \right) \left( \frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 = (-2)(-2) - 0^2 = 4 \]
Since \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the critical point \((3, 2)\) is identified as a relative maximum. The negative value of \( \frac{\partial^2 f}{\partial x^2} \) confirms that the function curves downward at this point.

Partial derivatives are a fundamental tool in multivariable calculus used to analyze functions of several variables. A partial derivative measures how a function changes as only one variable changes, while all others remain constant.
In our exercise, the function \( f(x, y) = 4y + 6x - x^2 - y^2 \) has two variables, \( x \) and \( y \). To find the critical points, we first take the partial derivatives with respect to \( x \) and \( y \). This is achieved by differentiating the function while treating the other variable as a constant:

• For \( x \), this results in \( \frac{\partial f}{\partial x} = 6 - 2x \).
• For \( y \), this results in \( \frac{\partial f}{\partial y} = 4 - 2y \).

Setting these derivatives to zero helps in discovering critical points where the function does not change instantaneously in any direction. Understanding partial derivatives is crucial for analyzing how a function behaves and for pinpointing critical points in the context of optimizing and graphing functions in higher dimensions.