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 𝑥 and 𝑦. Then, we can express the fact that
their sum is 156 as 𝑥 plus 𝑦 equals 156. We want to minimize the sum of
their squares, which we can call 𝑠. 𝑠 is equal to 𝑥 squared plus 𝑦
squared. In order to do so, we need to find
the values of 𝑥 and 𝑦 for which the rate of change of 𝑠 with respect to either 𝑥
or 𝑦, is equal to zero. This means that we need to
differentiate 𝑠 with respect to either 𝑥 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 𝑥 and then
substitute this expression for 𝑦 into our equation for 𝑠 to give an equation in
terms of 𝑥 only. Distributing the parentheses and
then simplifying gives 𝑠 is equal to two 𝑥 squared minus 312𝑥 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 𝑥, that’s d𝑠 by d𝑥, is equal to zero. We can use the power rule to find
this derivative. And we see that d𝑠 by d𝑥 is equal
to four 𝑥 minus 312. We then set this derivative equal
to zero and solve for 𝑥. We first add 312 to each side and
then divide by four, giving 𝑥 equals 78.
So, we found the value of 𝑥 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 𝑥 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 𝑥. Differentiating d𝑠 by d𝑥 again
gives d two 𝑠 by d𝑥 squared is equal to four. The second derivative of 𝑠 with
respect to 𝑥 is, therefore, constant for all values of 𝑥. 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.