Question

The sum of two positive numbers is 16. What is the optimum value (maximum or minimum) for the sum of their squares?

Answer 1

$x=8$ and $y=8$ are the two positive integers whose sum is $16$ and sum of their squares is minimum.

What is optimum value ?

The optimum value is a minimum or maximum value of the objective function over the feasible region of an optimization problem.

If a function is strictly increasing in a definite interval and increases up to a fixed value and after this, it starts decreasing, then that point is called maximum point of the function and value of function at that point is called maximum value.

If a function is strictly decreasing in a definite interval and decreases up to a fixed value and after this, it starts increasing, then that points is called minimum point of the function and the value of function at that point is called minimum value.

Conditions for finding maxima and minima

The conditions for maxima and minima for a function $y=f(x)$ at a point $x=a$  are as follow:

1. Necessary condition

for maxima and minima, the necessary condition is

$f'(x)=\frac{dy}{dx}$

2.Suffiecient condition

for maxima and minima, the  necessary condition are

for maximum value

at $x=a,\frac{d^2y}{dx^2}$ should be negative.

for minimum value

at $x=a, \frac{d^2y}{dx^2}$ should be positive.

The sum of two positive number is $16$.

We have to find the maximum and minimum value for the sum of their squares.

The sum of two positive number is 16.

let the number be $x$ and $y$, such that $x > 0$ and $y > 0$

sum of the number is $x+y=16$

sum of squares of the number $S=x^2+y^2$

$x+y=16\\y=16-x ----------1\\S=x^2+y^2\\S=x^2+(16-x)^2-----------2$after substituting the value of y from equation 1

for finding the maximum and minimum of given function we can find it by differentiating the function with $x$ equal it to $0$

Differentiate the equation 2

$\frac{dS}{dx} =\frac{d}{dx}[x^2+(16-x)^2]\\\frac{dS}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}(16-x^2)\\ \frac{dS}{dx}=2x+2(16-x)(-1)---------3$

Now equating the first derivative equal to zero

so, $\frac{dS}{dx}=0$

$2x+2(16-x)(-1)=0\\2x-2(16-x)=0\\2x-32+2x=0\\4x-32=0\\4x=32\\x=\frac{32}{4}=8$

As $x > 0, x=8$

Now, for checking if the value of $S$ is minimum or maximum at $x=8$, we will perform the second derivative of $S$ with respect to $x$

$\frac{d^2S}{dx^2}=\frac{d}{dx}[2x+2(16-x)(-1)]\\\frac{d^2S}{dx^2}=\frac{d}{dx}[2x-2(16-x)]\\\frac{d^2S}{dx^2}=\frac{d}{dx}(2x)-2\frac{d}{dx}(16-x)\\\frac{d^2S}{dx^2}=2-2(0-1)\\\frac{d^2S}{dx^2}=2-0+2=4\\\frac{d^2S}{dx^2}=4$

According to the sufficient condition if the second derivative is positive then the value is minimum

hence for $x=8$ will be the minimum point of the function $S$.

Therefore the function $S$ sum of squares of the two number is minimum at $x=8$

from equation $1$

$y=16-x\\y=16-8\\y=8$

Therefore , $x=8$ and $y=8$ are the two positive numbers whose sum is $16$ and the sum of their squares is minimum.

Learn more about the optimum value (maximum or minimum) here

brainly.com/question/28284783

#SPJ4