Question

find two positive numbers, whose sum is 15 and sum of whose squares is minimum .​

Answer 1

Answer: 15/2 and 15/2

Step-by-step explanation:

Let that two positive integers be x and y.

Given that,

Sum of numbers = 15

x + y = 15

y = 15 - x  ................i)

Let the sum of squares be p

x² + y² = p

p = x² + y²

Putting y = 15 - x;

p = x² + (15 - x)²

p = x² + 15² + x² - 30x

Now, Differentiating both sides with respect to x,

$\frac{dp}{dx}=2x+0+2x-30\\\;\\\;\\\frac{dp}{dx}=4x-30\;\;\;........ii)$

For Maximum and Minimum values,

$\frac{dp}{dx}=0\\\;\\4x-30=0\\\;\\x=\frac{30}{4}\\\;\\x=\frac{15}{2}$

Differentiating equation ii) again,

$\frac{d^2p}{dx^2}=4\\\;\\\frac{d^2p}{dx^2}_{(x=\frac{15}{2})}=4\;\;\;\;\;\;\text{(Positive)}$

Hence, at x = 15/2, p is minimum.

Putting x = 15/2 in equation i)

y = 15 - 15/2

y = (30-15)/2

y = 15/2

Hence both numbers are same for this condition.

Answer 2

SOLUTION:-

Let the number be x

Given,

Sum of two positive number is 15

1st number + 2nd number= 15

=) x + 2nd number= 15

=) 2nd number= 15 - x...........(1)

Let S(x) be the sum of the squares of the numbers

=) S(x) = (1st number)² + (2nd number)²

=) S(x)= x² + (15-x)² [from (1)]

We need to minimize S(x)

Finding S'(x)

S(x) = x² + (15-x)²

$S'(x) = \frac{d( {x}^{2}15 - x) {}^{2} }{dx} \\ \\ = > \frac{d( {x}^{2}) }{dx} + \frac{d(15 - x) {}^{2} }{dx} \\ \\ = > 2x + 2(15 - x)( - 1) \\ = > 2x - 2(15 - x) \\ = > 2x - 30 + 2x \\ = > 4x - 30$

Putting S'(x)= 0

$= > 4x - 30 = 0 \\ = > 4x = 30 \\ = > x = \frac{30}{4} = \frac{15}{2} \\ \\ = > x = \frac{15}{2}$

Finding S''(x)

S'(x)= 4x-30

$S''(x) = \frac{d(4x - 30)}{dx} \\ \\ = > 4$

Putting x= 15/2 in S''(x)

=) S''(15/2)= 4 at x= 15/2

=) S'' (x)> 0 at x= 15/2

Therefore,

x= 15/2 is local minima

Thus,

S(x) is minimum at x= 15/2

Hence,

1st number, x= 15/2

2nd number, 15-x = 15-15/2= 30-15/2

=) 15/2

Hope it helps ☺️