Question

Find three numbers whose sum is $21$ snd whose sum of squares is a minimum. The three numbers are :

Answer 1

Step-1: Formation of required function:

Let the positive numbers be $\text{x, y and z.}$

Given that the sum of numbers is $21$

$$\begin{gathered} \Rightarrow x+y+z=21---\left(1\right) \\ \Rightarrow z=21-x-y \end{gathered}$$

Let the sum of the squares of the number be S.

$$\begin{gathered} \therefore S=x^2+y^2+z^2 \\ \Rightarrow S=x^2+y^2+{(12-x-y)}^2---\left(2\right) \end{gathered}$$

Step-2: Find critical points:

To find the minimum value of $S$, we will optimize the function and differentiate it partially with respect to $\text{x and y}$ and make $\frac{dS}{dx}=0$

Partially differentiating (2) w.r.t $x$

$$\begin{gathered} 2x+2(21-x-y)(-1)=0 \\ \Rightarrow 2x-42+2x+2y=0 \\ \Rightarrow 4x+2y-42=0 \\ \Rightarrow y=21-2x---\left(3\right) \end{gathered}$$

Partially differentiating equation (2) w.r.t $y$,

$$\begin{gathered} 2y+2(21-x-y)(-1)=0 \\ \Rightarrow 2y-42+2x+2y=0 \\ \Rightarrow 4y+2x-42=0 \\ \Rightarrow x=21-2y---\left(4\right) \end{gathered}$$

By substituting the value of $y=21-2x$in equation (4),

$$\begin{gathered} \Rightarrow x=21-2(21-2x) \\ \Rightarrow x=21-42+4x \\ \Rightarrow 4x-x=42-21 \\ \Rightarrow 3x=21 \\ \Rightarrow x=7 \end{gathered}$$

By substituting $x=7$in equation (3)

$$\begin{gathered} y=21-2\left(7\right) \\ \Rightarrow y=21-14 \\ \Rightarrow y=7 \end{gathered}$$

By substituting the values of $x=7andy=7$ in equation (1), we get,

$$\begin{gathered} 7+7+z=21 \\ \Rightarrow z=21-14=7 \end{gathered}$$

From, we get

$$\begin{gathered} S=x^2+y^2+z^2 \\ \Rightarrow S=7^2+7^2+7^2 \\ \Rightarrow S=147 \end{gathered}$$

Hence, the three numbers are $7,7,7$.