WEBVTT
00:00:01.248 --> 00:00:07.348
Find two numbers whose sum is 156 and the sum of whose squares is the least possible.
00:00:08.838 --> 00:00:13.408
Letβs allow these two numbers, which we donβt yet know, to be π₯ and π¦.
00:00:14.638 --> 00:00:21.278
Then, we can express the fact that their sum is 156 as π₯ plus π¦ equals 156.
00:00:22.758 --> 00:00:27.068
We want to minimize the sum of their squares, which we can call π .
00:00:27.188 --> 00:00:29.728
π is equal to π₯ squared plus π¦ squared.
00:00:30.028 --> 00:00:38.528
In order to do so, we need to find the values of π₯ and π¦ for which the rate of change of π with respect to either π₯ or π¦, is equal to zero.
00:00:40.028 --> 00:00:44.568
This means that we need to differentiate π with respect to either π₯ or π¦.
00:00:44.828 --> 00:00:48.698
First, though, we need to write π in terms of one variable only.
00:00:49.798 --> 00:00:52.318
The choice is entirely arbitrary in this problem.
00:00:52.698 --> 00:01:06.438
We could perform a simple rearrangement of our first equation to give π¦ equals 156 minus π₯ and then substitute this expression for π¦ into our equation for π to give an equation in terms of π₯ only.
00:01:07.758 --> 00:01:18.608
Distributing the parentheses and then simplifying gives π is equal to two π₯ squared minus 312π₯ plus 24336.
00:01:19.918 --> 00:01:26.058
Remember, weβre looking to minimize this sum of squares, so we need to find the critical points of π .
00:01:26.248 --> 00:01:33.048
To do so, we need to find where the first derivative of π with respect to π₯, thatβs dπ by dπ₯, is equal to zero.
00:01:34.348 --> 00:01:36.768
We can use the power rule to find this derivative.
00:01:36.768 --> 00:01:41.398
And we see that dπ by dπ₯ is equal to four π₯ minus 312.
00:01:42.668 --> 00:01:46.558
We then set this derivative equal to zero and solve for π₯.
00:01:47.618 --> 00:01:53.618
We first add 312 to each side and then divide by four, giving π₯ equals 78.
00:01:55.078 --> 00:01:58.648
So, we found the value of π₯ at which π has a critical point.
00:01:58.798 --> 00:02:00.598
But there are two things we need to do.
00:02:00.958 --> 00:02:04.078
In a moment, weβll confirm that this is indeed a minimum.
00:02:04.338 --> 00:02:11.518
But first, we also need to find the value of π¦, which we can do by substituting the value of π₯ into our linear equation.
00:02:12.768 --> 00:02:18.198
We see that π¦ is equal to 156 minus 78, which is equal to 78.
00:02:19.718 --> 00:02:22.548
To confirm that this critical point is indeed a minimum.
00:02:22.548 --> 00:02:26.698
We need to find the second derivative of the function π with respect to π₯.
00:02:27.138 --> 00:02:32.578
Differentiating dπ by dπ₯ again gives d two π by dπ₯ squared is equal to four.
00:02:33.778 --> 00:02:38.638
The second derivative of π with respect to π₯ is, therefore, constant for all values of π₯.
00:02:38.758 --> 00:02:44.828
And more importantly, it is positive, which confirms that this critical point is indeed a minimum.
00:02:46.228 --> 00:02:53.728
The two numbers then whose sum is 156, which have the minimum sum of squares, are 78 and 78.