Video: Finding Two Numbers with the Minimum Sum of Their Squares given Their Sum Using Differentiation

Find two numbers whose sum is 156 and the sum of whose squares is the least possible.

02:55

Video Transcript

Find two numbers whose sum is 156 and the sum of whose squares is the least possible.

Let’s allow these two numbers, which we don’t yet know, to be π‘₯ and 𝑦. Then, we can express the fact that their sum is 156 as π‘₯ plus 𝑦 equals 156. We want to minimize the sum of their squares, which we can call 𝑠. 𝑠 is equal to π‘₯ squared plus 𝑦 squared. In order to do so, we need to find the values of π‘₯ and 𝑦 for which the rate of change of 𝑠 with respect to either π‘₯ or 𝑦, is equal to zero. This means that we need to differentiate 𝑠 with respect to either π‘₯ or 𝑦.

First, though, we need to write 𝑠 in terms of one variable only. The choice is entirely arbitrary in this problem. We could perform a simple rearrangement of our first equation to give 𝑦 equals 156 minus π‘₯ and then substitute this expression for 𝑦 into our equation for 𝑠 to give an equation in terms of π‘₯ only. Distributing the parentheses and then simplifying gives 𝑠 is equal to two π‘₯ squared minus 312π‘₯ plus 24336.

Remember, we’re looking to minimize this sum of squares, so we need to find the critical points of 𝑠. To do so, we need to find where the first derivative of 𝑠 with respect to π‘₯, that’s d𝑠 by dπ‘₯, is equal to zero. We can use the power rule to find this derivative. And we see that d𝑠 by dπ‘₯ is equal to four π‘₯ minus 312. We then set this derivative equal to zero and solve for π‘₯. We first add 312 to each side and then divide by four, giving π‘₯ equals 78.

So, we found the value of π‘₯ at which 𝑠 has a critical point. But there are two things we need to do. In a moment, we’ll confirm that this is indeed a minimum. But first, we also need to find the value of 𝑦, which we can do by substituting the value of π‘₯ into our linear equation. We see that 𝑦 is equal to 156 minus 78, which is equal to 78.

To confirm that this critical point is indeed a minimum. We need to find the second derivative of the function 𝑠 with respect to π‘₯. Differentiating d𝑠 by dπ‘₯ again gives d two 𝑠 by dπ‘₯ squared is equal to four. The second derivative of 𝑠 with respect to π‘₯ is, therefore, constant for all values of π‘₯. And more importantly, it is positive, which confirms that this critical point is indeed a minimum. The two numbers then whose sum is 156, which have the minimum sum of squares, are 78 and 78.

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