91Ó°ÊÓ

In multivariable calculus, finding critical points involves determining where a function's gradient is zero. The gradient of a function, denoted as \( abla f(x, y) \), is a vector containing its partial derivatives with respect to its variables. For our function, \( f(x, y) = 5y e^x - e^{5x} - y^5 \), this requires calculating:

• \( \frac{\partial f}{\partial x} = 5y e^x - 5e^{5x} \)
• \( \frac{\partial f}{\partial y} = 5e^x - 5y^4 \)

Setting these partial derivatives equal to zero helps us find potential critical points by solving the resulting system of equations. In this case, solving leads to the critical point \((x, y) = (0, 1)\).This point occurs where both partial derivatives equal zero, marking a spot on the surface of the function where it may change behavior — such as a maximum, minimum, or saddle point.

A key aspect of multivariable calculus is using the second derivative test to classify critical points as local maxima, minima, or saddle points. For the function examined here, the critical point \((0, 1)\) is analyzed for local behavior. This involves calculating second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} = 5y e^x - 25e^{5x} \)
• \( \frac{\partial^2 f}{\partial y^2} = -20y^3 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 5e^x \)

We then determine the Hessian matrix's determinant \( D \) at this point:\[ D = f_{xx} f_{yy} - (f_{xy})^2 = (-20)(-20) - (5)^2 = 375 \]A positive \( D > 0 \) together with a negative \( f_{xx} < 0 \) implies a local maximum at this critical point. Thus, \((0, 1)\) is a local maximum, indicating a peak in the function's graph at that location.

An unbounded function is one that doesn't have an upper or lower limit in its range. For functions like \( f(x, y) = 5y e^x - e^{5x} - y^5 \), analyzing their behavior along specific axes reveals if they are unbounded. Here, we specifically look at what happens as we let \( y \to \infty \) while keeping \( x = 0 \):\[ f(0, y) = 5y - 1 - y^5 \]When \( y \) becomes very large, the \( y^5 \) term grows much faster than the linear term \( 5y \), making the expression \( 5y - 1 - y^5 \) tend towards \(-\infty\). This implies that the function is unbounded below along the y-axis, illustrating that the function does not have a global maximum value.

Partial derivatives are fundamental tools in multivariable calculus. They measure how a function changes as one variable changes, keeping other variables constant. For example, in the function \( f(x, y) = 5y e^x - e^{5x} - y^5 \), the partial derivative with respect to \( x \):\[ \frac{\partial f}{\partial x} = 5y e^x - 5e^{5x} \]indicates how the function changes as \( x \) varies, with \( y \) held fixed. Similarly, the partial derivative with respect to \( y \):\[ \frac{\partial f}{\partial y} = 5e^x - 5y^4 \]shows how the function changes as \( y \) varies, with \( x \) fixed.

• Partial derivatives are used to find critical points by setting them to zero.
• Second partial derivatives help classify the nature of these critical points.

By understanding partial derivatives, we gain crucial insights into the behavior of multivariable functions.