Question

the sum of two positive numbers is 24 . find the numbers so that the sum of their squares is minimum ​

Answer 1

Answer:

12 and 12

Step-by-step explanation:

Let first number be $x$ and the second be 24-$x$

Let S be the sum of squares of these two numbers

So

S= $x^{2}$ + $(24-x)^2$

Using $(a-b)^2=a^2-2ab+b^2$

S= $2x^2-48x+576$

Now differentiating wrt $x$

$\frac{dS}{dt} = 4x-48$

For extreme values of S we have $\frac{dS}{dx}$=0

∴$4x-48=0\\x=12$

∴at $x$= 12 either there is maxima or minima

Again differentiating wrt $x$

$\frac{d^2S}{dx^2} = 4$

Substituting  x=12

$(\frac{d^2S}{dx^2})_{x=12} = 4$

4>0

Therefore S has minima at 12

And 24-12=12

So the two numbers are 12 and 12

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