Second derivative test. Worked example
Let’s do an example. Let’s say we look at the function f (x, y) = x + y + 1/xy where x and y are positive. We are looking at this function because we do not want the denominator to be 0, so we are just looking at that situation. So let’s look first for critical points (the minimum and maximum). To solve this problem, we need to compute the first derivatives of f(x, y). The derivative of one over x is negative over x2. Our goal is to set this expression equal 0 to 0 and have it equal 0 for all values of x. The derivative of one over xy is negative 1 over xy2. We want this expression to equal 0 for all values of y. So what are the equations we have to solve? The equation x squared y equals 1 is equivalent to x y squared equals 1. If we substitute one for the other in the first equation, we get x squared y equals 1. Let x equal one, and y equal one. By dividing each of these by their respective denominators, we arrive at the conclusion that one divided by one is also a solution. However, this is not the only solution, as any number divided by itself will result in a unit fraction. Therefore, we can rewrite this fraction as x/x = y/y = 1/1 = 1. If x equals y, then y cubed equals 1. If I put that into here. it will tell it me that y equals 1 1 and therefore x equals 1 as well. So there’s only one solution-(1,1). Example: f(x,y)=x+y+1/xy x,y>0 Min? Max? Critical points? f_x=1-1/x^2y=0 f_y=1-1/xy^2=0 {x^2y=1;xy^2=1} x=y y^3=1->y=1 x=1 Only one critical point: (1, 1)
