Courses
Courses for Kids
Free study material
Free LIVE classes
More

# Find the two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Last updated date: 23rd Mar 2023
Total views: 307.8k
Views today: 2.86k
Verified
307.8k+ views
Hint: We have to use the derivative tests to find the minimum value of a function. Here we are going to represent two numbers into a function of some variable x.

Let the two numbers be x, y.

From the given information, the sum of those two numbers is 15.
$\Rightarrow x + y = 15$ ... (1)
Let ‘S’ be the sum of squares of those numbers
$\Rightarrow S = {x^2} + {y^2}$

From equation (1) we can substitute y value, to make the above equation in terms of x.
$\Rightarrow S = {x^2} + {(15 - x)^2}$
‘S’ is a function of x, so we can represent it as
$\Rightarrow S(x) = {x^2} + {(15 - x)^2}$

We want the minimum value of S(x), so differentiating S(x) with respect to x
$\Rightarrow \dfrac{{d(S(x))}}{{dx}} = \dfrac{d}{{dx}}({x^2} + {(15 - x)^2})$
$\Rightarrow S\prime (x) = 2x + 2(15 - x)( - 1) = 2x - 30 + 2x$
$\Rightarrow S\prime (x) = 4x - 30$ .... (2)
For maximum or minimum value of S(x), making
$\Rightarrow 4x - 30 = 0$
$\Rightarrow x = \dfrac{{30}}{4} = 7.5$
So at x = 7.5 we get the minimum value of S(x).

Differentiating equation (2) with respect to x
$$\Rightarrow S\prime \prime (x) = \dfrac{d}{{dx}}\left( {4x - 30} \right)$$
$$\Rightarrow S\prime \prime (x) = 4$$
Substituting x=7.5 in $$S\prime \prime (x)$$
$$\Rightarrow S\prime \prime (7.5) = 4 > 0$$
At x= 7.5, $$S\prime \prime (x)$$is positive. Thus S’(x) is minimum at x=7.5
$$\therefore$$ The numbers are $$x = \dfrac{{15}}{2},y = 15 - x = \dfrac{{15}}{2}$$

Note: The first derivative is used to find the critical points. That if we take the derivative of a function and set it equal to zero and solve we find critical points. Critical points give the maximum or minimum value of the function. The second derivative used to find the possible point of inflection. If the second derivative of a function is positive then the graph will be increasing, if the second derivative is negative the graph is increasing. The place where the curve changes from either increasing to decreasing or vice versa is called an inflection point.