91Ó°ÊÓ

When finding critical points of a function of multiple variables, partial derivatives are a handy tool. Partial derivatives help us understand how the function changes concerning one particular variable while keeping the others constant.
To find the partial derivative of the function \(f(x, y)=x^{2}+x y-2 x-y+1\), we need to derive it partially with respect to \(x\) and \(y\). These derivations give us the function's slopes in the \(x\) and \(y\) directions.

• The partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), is \(2x + y - 2\).
• The partial derivative with respect to \(y\), denoted \(\frac{\partial f}{\partial y}\), is \(x - 1\).

These derivatives tell us how the function is changing at any point \((x, y)\). Finding where both are zero assists in locating critical points, where the function doesn't increase or decrease in any direction.

After calculating the partial derivatives, the next step towards finding critical points involves setting these derivatives to zero, which creates a system of equations.
In our exercise, we have two equations from the partial derivatives:

• \(2x + y - 2 = 0\)
• \(x - 1 = 0\)

This is a simple system of linear equations in two variables (\(x\) and \(y\)). Solving these equations simultaneously will give the points where the function potentially achieves a local maximum, minimum, or saddle point.
By solving this system, we find where both partial derivatives are zero, meaning the rate of change in either direction vanishes at that point. This framework walks us towards identifying potential critical points.

The substitution method is a classic technique for solving a system of equations, simplifying the process by substituting one equation into another.
In our exercise, we derived an equation from the partial derivative with respect to \(x\), \(x - 1 = 0\). It tells us directly that \(x = 1\).
Using substitution, we place this value of \(x\) into the first equation, \(2x + y - 2 = 0\). This becomes:
\[ 2(1) + y - 2 = 0 \]
Here, after simplifying, we find \(y = 0\). So, our solution is a single point \((x, y) = (1, 0)\).
The substitution method makes tackling these problems straightforward by turning a multi-variable problem into a single-variable problem. Once you find the value for one variable, solving for the second becomes much easier.