Functions of several variables: Critical points
Critical points
A differentiable function \(f(x,y)\) has a critical point at\((x,y)=(a,b)\) if the gradient \(\nabla f\) at that point is the zero vector. This means that all first partial derivatives at this point are zero, and the tangent plane of the graph if \(f\) at the stationary point \(\bigl(a,b,f(a,b)\bigr)\) is horizontal.
Ciritcal points can be found by solving the following system of equations: \[\left\{\begin{array}{c} f_x(x,y)=0\\ f_y(x,y)=0\end{array}\right.\]
There is one critical point: \((-2,9)\).
Find the critical points of the function \[f(x,y)=3x^2y+xy^2-3xy\] The first partial derivatives are \[f_x(x,y)=6xy+y^2-3y\qquad\text{and}\qquad 3x^2+2xy-3x\] The critical points are the solutions of the following system of equations \[\left\{\begin{aligned} 6xy+\phantom{2}y^2-3y&=0\\ 3x^2+2xy-3x&=0\end{aligned}\right.\] We divide this problem into subproblems by factorizing the left-hand sides of the equations: \[\left\{\begin{aligned} y(6x+\phantom{2}y-3)&=0\\ x(3x+2y-3)&=0\end{aligned}\right.\] We can obviously get each \(f_x(x,y)\) and \(f_y(x,y)\) equal to zero in two ways. Combination of the possibilities leads to four solutions:
- \(\left\{\begin{aligned} x&=0\\ y&=0\end{aligned}\right. \)
- \(\left\{\begin{aligned} 3x+2y-3&=0\\ y&=0\end{aligned} \right. \quad \implies \quad \) \(\left\{\begin{aligned} x&=1\\ y&=0\end{aligned} \right. \)
- \(\left\{\begin{aligned} x&=0\\ 6x+y-3&=0\end{aligned} \right. \quad \implies\) \(\quad \left\{ \begin{aligned} x&=0\\ y&=3\end{aligned} \right. \)
- \(\left\{\begin{aligned} 6x+\phantom{2}y-3&=0\\ 3x+2y-3&=0\end{aligned} \right. \quad \implies \) \(\quad\left\{\begin{aligned} -3y+3&=0\\ 3x+2y-3&=0\end{aligned}\right. \quad \implies\) \(\quad \left\{ \begin{aligned} x&=\frac{1}{3}\\ y&=1\end{aligned} \right. \)
There are four criticial points: \((0,0)\), \((1,0)\), \((0,3)\) and \((\tfrac{1}{3},1)\).
