Divide \[12\] into two parts such that the product of the square of one part and the fourth power of the second part is maximum. Then find the pair of numbers.
A. \[6,6\]
B. \[5,7\]
C. \[4,8\]
D. \[3,9\]

Hint: In the given question, we have to divide \[12\] into two parts such that the product of the square of one part and the fourth power of the second part is maximum. Consider the first and second parts then multiply the square of the first part and the fourth power of the second part. Differentiate this new function with respect to \[x\]. Equate the differential equation to 0 and find the local maxima or local minima of the function. Then substitute the values of \[x\] in the original equation, and we will find the pair of numbers.

Formula used:
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
When the first derivative of a function at a point \[x = a\] is 0, the point is either local minima or local maxima.

Complete step by step solution:
The given number is \[12\].
It is divided into two parts such that the product of the square of one part and the fourth power of the second part is maximum.
Let consider \[x\] and \[12 - x\] are the two parts. Where \[0 < x < 12\].
Then,
\[f\left( x \right) = {x^4}{\left( {12 - x} \right)^2}\] \[.....equation\left( 1 \right)\]
\[ \Rightarrow \]\[f\left( x \right) = {x^4}\left( {144 + {x^2} - 24x} \right)\]
\[ \Rightarrow \]\[f\left( x \right) = {x^6} - 24{x^5} + 144{x^4}\]
Now differentiate above equation with respect to \[x\].
\[f'\left( x \right) = 6{x^5} - 120{x^4} + 576{x^3}\]
To calculate the local maxima or minima equate above equation to zero.
\[6{x^5} - 120{x^4} + 576{x^3} = 0\]
\[ \Rightarrow \]\[6{x^3}\left( {{x^2} - 20x + 96} \right) = 0\]
\[ \Rightarrow \]\[6{x^3}\left( {{x^2} - 12x - 8x + 96} \right) = 0\]
\[ \Rightarrow \]\[(6{x^3}\left( {x\left( {x - 12} \right) - 8\left( {x - 12} \right)} \right) = 0\]
\[ \Rightarrow \]\[6{x^3}\left( {x - 12} \right)\left( {x - 8} \right) = 0\]
\[ \Rightarrow \] \[x = 0\], \[x - 12 = 0\], \[x - 8 = 0\]
\[ \Rightarrow \]\[x = 0\], \[x = 12\], \[x = 8\]

Now substitute the values of \[x\] in equation \[\left( 1 \right)\].
For \[x = 0\]:
\[f\left( 0 \right) = {0^4}{\left( {12 - 0} \right)^2}\]
\[ \Rightarrow \]\[f\left( 0 \right) = 0\]

For \[x = 12\]:
\[f\left( {12} \right) = {\left( {12} \right)^4}{\left( {12 - 12} \right)^2}\]
\[ \Rightarrow \]\[f\left( {12} \right) = {\left( {12} \right)^4}{\left( 0 \right)^2}\]
\[ \Rightarrow \]\[f\left( {12} \right) = 0\]

For \[x = 8\]:
\[f\left( 8 \right) = {\left( 8 \right)^4}{\left( {12 - 8} \right)^2}\]
\[ \Rightarrow \]\[f\left( 8 \right) = \left( {4096} \right){\left( 4 \right)^2}\]
\[ \Rightarrow \]\[f\left( 8 \right) = \left( {4096} \right)\left( {16} \right)\]
\[ \Rightarrow \]\[f\left( 8 \right) = 65536\]
\[f\left( x \right)\] has a maximum when \[x = 8\].
So,
\[12 - x = 12 - 8\]
\[ \Rightarrow \]\[12 - x = 4\]
Hence the correct option is option C.

Note: We can use the second order derivative to find the maxima for the function \[f\left( x \right)\]. If \[f''\left( x \right) < 0\] at \[x = a\] then the function is maxima at \[x = a\]. If \[f''\left( x \right) > 0\] at \[x = a\] then the function is minima at \[x = a\]. But if \[f''\left( x \right) = 0\] at \[x = a\] then \[x = a\] is an inflection point.