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Question: The sum of squares of two parts of a number 100 is minimum, then two parts are: \( \left( a \r...

The sum of squares of two parts of a number 100 is minimum, then two parts are:
(a)50,50 (b)25,75 (c)40,60 (d)30,70  \left( a \right)50,50 \\\ \left( b \right)25,75 \\\ \left( c \right)40,60 \\\ \left( d \right)30,70 \\\

Explanation

Solution

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

Complete step-by-step answer:
Let xx and yy be two parts of 100.
So, we can write as x+y=100x + y = 100
y=100x\Rightarrow y = 100 - x
So, xx and 100x100 - x are two parts of 100 .
Now, according to question
f(x)=(x)2+(100x)2............(1)f\left( x \right) = {\left( x \right)^2} + {\left( {100 - x} \right)^2}............\left( 1 \right)
For maxima and minima, f(x)=dfdx=0f'\left( x \right) = \frac{{df}}{{dx}} = 0 .
So, Differentiate (1) equation with respect to x .

f(x)=ddx((x)2+(100x)2) f(x)=2x+2(100x)(1) f(x)=4x200..........(2) f(x)=0 4x200=0 4x=200 x=50  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(x)=ddx(4x200) f(x)=4  f''\left( x \right) = \frac{d}{{dx}}\left( {4x - 200} \right) \\\ \Rightarrow f''\left( x \right) = 4 \\\
f(x)>0f''\left( x \right) > 0 for all value of x .
Now, f(x)>0f''\left( x \right) > 0 for x=50
So, x=50 is a minimum point.
Hence the function f(x)=(x)2+(100x)2f\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.