00:03
We have two real and positive numbers, x and y, such that they fulfill the following relationship, x squared times y equals 20.
00:18
Then we want to find the value of x that minimizes the expression x square plus 3xy.
00:27
So initially we have this function of two variables, f of xy, equals x squared plus 3xy, x y so it will be a minimization problem of two variables initially but we know that x and y have this relationship given by this equation x squared times y equals 20 and from here we can reduce this function to one variable the variable x only because we can solve this equation here for y for example we can do that because x and y are both positive so x square is not zero so we can divide by x squared to the right so y equals 20 over x square and so we can replace this suppression here into the formula of f and find a function of one variable we call it f again and is dependent only on x and that's equal to x square plus 3x times y that is 20 over x square and that simplifies to x square plus 60 over x that is the function we want to study is f of x equal x square plus 60 over x so the minimization problem reduced to this one here of one variable.
02:34
So the set where we want to find this minimum value is an open set.
02:44
So we are going to see if we have only one critical point of this function.
02:50
That case the second derivative test can give us the answer.
02:57
Let's see the derivative of f.
03:00
Is equal to 2x minus 60 over x square that's the derivative of f and this can be written if we want as 2x cube minus 60 over x square and that's equal to two two common factor of x cube minus 30 over x squared square.
03:43
From here we identify that the derivative is zero, that is, to find the critical points, we solved this equation, and this derivative is zero if and only if the factor in the numerator x -cube minus 30 is zero.
04:05
And that happens if and only if x -cube is 30, and that happens if and only if being that is a an odd power here we have only one solution which is cubic root of 30 so this is the critical the only critical point of the function f of x and if we calculate the second derivative of f now we get 2 plus 120 over x cube and if we see substitute this critical point into the second derivative, we get a positive value because we will put here x -cube is 30 and all the numbers are positive, so the result is positive.
05:23
And being positive the second derivative, we know that we have a minimum value at x equal cubic root of 30.
05:30
Although the thing we can do, we can say is that the first derivative is, the first derivative is, 2 x cubed minus 30 over x square let's put first this here f has a global in fact i got to say correctly this implies that f f has a local minimum at cubic root 30 but being the only critical point in that open set for the positive x and the second derivative being positive then we can state that this global on the set of the positive x so that's why i'm putting the conclusion here directly and not writing all the thing but there are two parts to arrive to this conclusion global minimum at x equal cubic root of 30 but we can do another thing to be sure of that and that's an the first derivative the first derivative of f is let's remember 2 x cubed minus 30 over x squared this denominator here is positive always positive for x different from zero x is different from zero because x is positive and the numerator the sign of this constant is positive so the sign of all the expression depends on the sign of this factor x cubed minus 30 and it is clear that if x is greater than the cubic root of 30 then implies that the cube of x is greater than 30, and so the cubic x -cube minus 30 is positive.
07:45
And similarly, if x is less than cubic root of 30, then the cube of x is less than 30.
07:55
And so the cubic root of the x cubed minus 30 is negative that is because we do this step here because the function x to the third that is the third power is an increasing always increasing function and then we can say that the factor and so the first derivative changes sign when we cross the point cubic root of 30 it means that all that means that f is decreasing that is derivative negative for x less than cubic root of 30 that is decreasing on 0 cubic root of 30 that is x less than cubic root of 30 and we know x is positive so in this interval the function is decreasing and f is increasing on cubic root 30 plus infinite...