Question

Find the number whose sum is $16$ and the sum of whose square is minimum.

Answer 1

To find the number, first of all we will assume the numbers and then will use the condition that the sum of squares of the number is minimum. We will consider it a function of $y$and will differentiate it two times with respect to $x$. If the obtained value is positive, the function will be minimum. So, we will use the property that $\dfrac{dy}{dx}=0$ and will simplify to get the value of the number.

Complete step by step solution:
Let’s consider that the first number is $a$ and the second number is $\left( 16-a \right)$.
Now, from the given condition in the question:
$\Rightarrow {{a}^{2}}+{{\left( 16-a \right)}^{2}}$
We will assume the above equation as:
$\Rightarrow y={{a}^{2}}+{{\left( 16-a \right)}^{2}}$
Now, we will differentiate the above equation with respect to $x$ and will get $2a$ and $2\left( 16-a \right)\left( -1 \right)$as:
$\Rightarrow \dfrac{dy}{dx}=2a+2\left( 16-a \right)\left( -1 \right)$
Here, simplify the above equation. First multiply with $2$ in the bracketed terms as:
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=2a+\left( 2\times 16-2a \right)\left( -1 \right) \\\ & \Rightarrow \dfrac{dy}{dx}=2a+\left( 32-2a \right)\left( -1 \right) \\\ \end{aligned}$
Now, we will multiply with $\left( -1 \right)$ in the bracketed terms as:
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=2a+\left( 16\times \left( -1 \right)-2a\times \left( -1 \right) \right) \\\ & \Rightarrow \dfrac{dy}{dx}=2a+\left( -32+2a \right) \\\ \end{aligned}$
Here, we can open the bracket to simplify the above step as:
$\Rightarrow \dfrac{dy}{dx}=2a-32+2a$
Now, after adding $2a$ and $2a$, we will get $4a$ as:
$\Rightarrow \dfrac{dy}{dx}=4a-32$
As we get the value of $\dfrac{dy}{dx}$. We will again differentiate it with respect to $x$ and will get $4$ as:
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4$
Since, the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is greater than $0$that means the value of equation should be minimum. So, we will use the property as:
$\Rightarrow \dfrac{dy}{dx}=0$
Here, we will substitute the obtained value $4a-32$ for $\dfrac{dy}{dx}$ in the above property as:
$\Rightarrow 4a-32=0$
Now, we will add $32$ both sides in the above step and will solve it as:
$\begin{aligned} & \Rightarrow 4a-32+32=0+32 \\\ & \Rightarrow 4a=32 \\\ \end{aligned}$
Here, we will divide by $4$ both sides as:
$\Rightarrow \dfrac{4a}{4}=\dfrac{32}{4}$
After simplifying it, we will have:
$\Rightarrow a=8$
Here, we got the first number. Then, the second number is
$\Rightarrow 16-8=8$
Hence, the both numbers are $8$ and $8$.

Note: We can check that the solution is correct or not in the following way as:
Since, the sum of the numbers is $16$. We check all the possible combination of two numbers that are:
$\Rightarrow \left( 1,15 \right),\left( 2,14 \right),\left( 3,13 \right),\left( 4,12 \right)\left( 5,11 \right)\left( 6,10 \right)\left( 7,9 \right),\left( 8,8 \right)$
Now, we will calculate the square of these combinations as:
$\Rightarrow \left( 1,225 \right),\left( 4,196 \right),\left( 9,169 \right),\left( 16,144 \right),\left( 25,121 \right),\left( 36,100 \right),\left( 49,81 \right),\left( 64,64 \right)$
Here, do the addition of both numbers and check whose sum is smallest.
$=226,200,178,160,146,136,130,128$
It can be seen that the sum of squares of $8$ and $8$ gives a minimum value.
Hence, the solution is correct.