# The sum of squares of two parts of a number 100 is minimum, then two parts are:

$

\left( a \right)50,50 \\

\left( b \right)25,75 \\

\left( c \right)40,60 \\

\left( d \right)30,70 \\

$

Last updated date: 16th Mar 2023

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Answer

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Hint: Use application of derivative to find maxima and minima .For maximum and minimum point derivative of function $f'\left( x \right) = \frac{{df}}{{dx}} = 0$ and for check maxima and minima use second derivative test, $f''\left( x \right) > 0$ minima point and $f''\left( x \right) < 0$ maxima point.

Complete step-by-step answer:

Let $x$ and $y$ be two parts of 100.

So, we can write as $x + y = 100$

$ \Rightarrow y = 100 - x$

So, $x$ and $100 - x$ are two parts of 100 .

Now, according to question

$f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}............\left( 1 \right)$

For maxima and minima, $f'\left( x \right) = \frac{{df}}{{dx}} = 0$ .

So, Differentiate (1) equation with respect to x .

\[

f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( x \right)}^2} + {{\left( {100 - x} \right)}^2}} \right) \\

\Rightarrow f'\left( x \right) = 2x + 2\left( {100 - x} \right)\left( { - 1} \right) \\

\Rightarrow f'\left( x \right) = 4x - 200..........\left( 2 \right) \\

f'\left( x \right) = 0 \\

\Rightarrow 4x - 200 = 0 \\

\Rightarrow 4x = 200 \\

\Rightarrow x = 50 \\

\]

Now, use the second derivative test for check x=50 is a maxima or minima point .

So, Differentiate (2) equation with respect to x .

$

f''\left( x \right) = \frac{d}{{dx}}\left( {4x - 200} \right) \\

\Rightarrow f''\left( x \right) = 4 \\

$

$f''\left( x \right) > 0$ for all value of x .

Now, $f''\left( x \right) > 0$ for x=50

So, x=50 is a minimum point.

Hence the function $f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}$ minimum at x=50 .

So, the required parts are 50 and 50 .

So, the correct option is (a).

Note: Whenever we face such types of problems we use some important points. First we assume the parts of a number and make a function in one variable according to the question then differentiate the function for maxima and minima then use a second derivative test to confirm the point is maxima or minima.

Complete step-by-step answer:

Let $x$ and $y$ be two parts of 100.

So, we can write as $x + y = 100$

$ \Rightarrow y = 100 - x$

So, $x$ and $100 - x$ are two parts of 100 .

Now, according to question

$f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}............\left( 1 \right)$

For maxima and minima, $f'\left( x \right) = \frac{{df}}{{dx}} = 0$ .

So, Differentiate (1) equation with respect to x .

\[

f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left( x \right)}^2} + {{\left( {100 - x} \right)}^2}} \right) \\

\Rightarrow f'\left( x \right) = 2x + 2\left( {100 - x} \right)\left( { - 1} \right) \\

\Rightarrow f'\left( x \right) = 4x - 200..........\left( 2 \right) \\

f'\left( x \right) = 0 \\

\Rightarrow 4x - 200 = 0 \\

\Rightarrow 4x = 200 \\

\Rightarrow x = 50 \\

\]

Now, use the second derivative test for check x=50 is a maxima or minima point .

So, Differentiate (2) equation with respect to x .

$

f''\left( x \right) = \frac{d}{{dx}}\left( {4x - 200} \right) \\

\Rightarrow f''\left( x \right) = 4 \\

$

$f''\left( x \right) > 0$ for all value of x .

Now, $f''\left( x \right) > 0$ for x=50

So, x=50 is a minimum point.

Hence the function $f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}$ minimum at x=50 .

So, the required parts are 50 and 50 .

So, the correct option is (a).

Note: Whenever we face such types of problems we use some important points. First we assume the parts of a number and make a function in one variable according to the question then differentiate the function for maxima and minima then use a second derivative test to confirm the point is maxima or minima.

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