00:01
In this question, we want to look at any critical points within this function, first without any restrictions on x or y, and followed by some restrictions.
00:13
So our first set would be to find the derivative of this function with respect to t.
00:18
That means that we are solving this parametrically.
00:22
So we want to note that df d t is equals to df the x times dx d t plus d fd y, plus d fd y, times the y d t we find the derivatives individually and then multiply and add them up we then set the derivative of f with respect to t to zero and try to solve it for the value of t where there is a critical point so the value of t here that we found is 4 divided by 5 so in the course of finding t we do have to substitute x and y for its parametric equations and with this value of t we can find the values of x and y at this value of t we then find the value of the function at this set of x and y values and we can conclude that it is a minimum point because when we look at the equation and if we were to substitute the parametric equations into x and y, we see that it is an equation with a minimum point rather than a maximum point.
01:36
So now we would like to restrict our range of t from 0 to 1.
01:43
The first critical point we found is still relevant, but let us look at the end points as well to see if they might be low.
01:53
Or they might be lower than the minimum point we found earlier.
02:00
So our first point of interest would be when t is equal to 0, we find the corresponding x and y values, and then we find the value of the function at this set of x and y values...