Question

Divide 12 into two parts so that the product of the square of one part and the fourth power of the other is the maximum.

Answer 1

Hint: In this question, we shall first take two numbers in x as variables and then write a function in that variable (say, x). Then, we shall write an equation for the condition given in the question and then evaluate its point of change using the first derivative. At that point of change, we shall verify the condition for maxima using the second derivative. The value of x which satisfies the condition $f''(x)<0$ will give us the answer.

Complete step-by-step solution
It is given that 12 is divided into two parts. Let us take the first part to be $x$. Then, we will get the second part as $12-x$.
We have to find the square of one part, fourth power of the other part and then take their product. So, according to the question, we can write the expression as:
$\begin{align}
  & f(x)={{x}^{4}}{{\left( 12-x \right)}^{2}} \\
 & f(x)={{x}^{4}}\left( 144+{{x}^{2}}-24x \right) \\
 & f(x)={{x}^{6}}-24{{x}^{5}}+144{{x}^{4}} \\
\end{align}$
We have the function now. So, to find the maximum value, we will first find the first derivative, equate it to zero. Then, we will get a critical point. Next, we will find the second derivative at that critical point and check if it is < 0 or not to satisfy the maximum condition.
We can find the derivative of $f(x)$ by differentiating it w.r.t. x, as:
$f'(x)=\dfrac{d}{dx}\left( {{x}^{6}}-24{{x}^{5}}+144{{x}^{4}} \right)$
We will use the fact that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and solve as below,
$\begin{align}
  & f'(x)=6{{x}^{5}}-24\times 5{{x}^{4}}+144\times 4{{x}^{3}} \\
 & f'(x)=6{{x}^{5}}-120{{x}^{4}}+576{{x}^{3}} \\
\end{align}$
For $f(x)$ to be maximum or minimum, we will equate $f'(x)=0$
Hence, we will substitute the value of $f'(x)$ in the above condition, as:
$\begin{align}
  & 6{{x}^{5}}-120{{x}^{4}}+576{{x}^{3}}=0 \\
 & \Rightarrow 6{{x}^{3}}\left( {{x}^{2}}-20x+96 \right)=0 \\
 & \Rightarrow {{x}^{2}}-20x+96=0 \\
 & \Rightarrow x\left( x-12 \right)-8\left( x-12 \right)=0 \\
 & \Rightarrow \left( x-12 \right)\left( x-8 \right)=0 \\
 & \Rightarrow x=8 \\
\end{align}$
Here, the value x = 12 is rejected as it will make the other part zero, i.e 12-x as zero. The value of x = 0 has also been rejected as it will result in one part being 12, which is not possible as per the condition in the question.
Now, to find if $f(x)$ is maximum at $x=8,$ we will check if $f''(x)<0.$
Hence,
$\begin{align}
  & f''\left( x \right)=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( {{x}^{6}}-24{{x}^{5}}+144{{x}^{4}} \right) \\
 & f''\left( x \right)=30{{x}^{4}}-480{{x}^{3}}+1728{{x}^{2}} \\
\end{align}$
Substituting the value of x = 8 in the above equation, we get:
$\begin{align}
  & f''\left( x \right)=30\times {{8}^{4}}-480\times {{8}^{3}}+1728\times {{8}^{2}} \\
 & f''\left( x \right)={{8}^{2}}\left( 30\times {{8}^{2}}-480\times 8+1728 \right) \\
 & f''\left( x \right)=64\left( 1920-3840+1728 \right) \\
 & f''\left( x \right)=64\times -192 \\
 & f''\left( x \right)=-12288 \\
\end{align}$
We can see that $f''(x)<0$
Hence,we have got that $f(x)$ is maximum for x = 8.
Therefore, the first part will be x $\Rightarrow 8$.
The second part will be $(12-x)\Rightarrow 12-8\Rightarrow 4$.
The two parts are 8 and 4 for which the conditions in the question hold true.

Note: The roots of the equation $f'(x)=0$ will be five in number, since it is a 5-degree equation in x. The roots of the equation are 0, 0, 0, 8, and 12. However, this is beyond the scope of the question. So, students must focus on getting that one value of x which satisfies all the conditions as the critical point. Always perform the second derivative test and do not stop after the first derivative to complete such types of questions.