Question

Divide $20$ into two parts such that the product of one part and the cube of the other is maximum.
(A) $13$ and $7$
(B) $14$ and $6$
(C) $15$ and $5$
(D) $16$ and $4$

Answer 1

Hint : In the given problem, first we will assume that $x$ and $y$ are two parts of a given number. Then, we will write the product of one part and the cube of the other part. We will consider a function $f\left( x \right)$ or $f\left( y \right)$ after writing the product in terms of only $x$ or only $y$. Then, we will use a second derivative test to find the maximum value of that function.

Complete step-by-step answer :
In the given problem, we need to find two parts of the number $20$ such that the product of one part and the cube of the other part is maximum. For this, let us assume that $x$ and $y$ are two parts of the number $20$. So, we can write $x + y = 20 \cdots \cdots \left( 1 \right)$. Now the product of one part $x$ and the cube of the other part $y$ can be written as $x{y^3}$. That is, product $= x{y^3} \cdots \cdots \left( 2 \right)$. From $\left( 1 \right)$, we can write $x = 20 - y$. Substitute $x = 20 - y$ in $\left( 2 \right)$, we get product$= \left( {20 - y} \right){y^3}$
$\Rightarrow$ Product$= 20{y^3} - {y^4}$
Now we have to find the maximum value of this product. Note that here $y$ is only variable in this product. Let us say $f\left( y \right) = 20{y^3} - {y^4}$. To find the maximum value of $f\left( y \right)$, we will use a second derivative test. Let us find the derivative of $f\left( y \right)$ with respect to $y$. So, we get
$f'\left( y \right) = 60{y^2} - 4{y^3}$. Note that here we used the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ for differentiation.
Let us consider $f'\left( y \right) = 0$ and solve it. So, we get
$60{y^2} - 4{y^3} = 0$
$\Rightarrow 4{y^2}\left( {15 - y} \right) = 0$
$\Rightarrow {y^2}\left( {15 - y} \right) = 0$
$\Rightarrow {y^2} = 0$ or $15 - y = 0$
$\Rightarrow y = 0$ or $y = 15$
Now we are going to find the second derivative of $f\left( y \right)$. That is, $f''\left( y \right) = 120y - 12{y^2}$.
Now we will find the value of $f''\left( y \right)$ at $y = 0$ and $y = 15$. Note that if $f''\left( y \right) < 0$ then we can say that $f\left( y \right)$ is maximum at that point.
For $y = 0$, $f''\left( y \right) = f''\left( 0 \right) = 120\left( 0 \right) - 12\left( 0 \right) = 0$
For $y = 15$, $f''\left( y \right) = f''\left( {15} \right) = 120\left( {15} \right) - 12\left( {{{15}^2}} \right) = - 900 < 0$
Here we can see that $f''\left( y \right) = - 900 < 0$ at $y = 15$. So, we can say that $f\left( y \right)$ is maximum at $y = 15$. Let us find the value of $x$ from $\left( 1 \right)$ by putting $y = 15$ in equation $\left( 1 \right)$. So, we get
$x + 15 = 20$
$\Rightarrow x = 20 - 15$
$\Rightarrow x = 5$
Hence, the required parts of the number $20$ are $5$ and $15$.

So, the correct answer is “Option C”.

Note : The second derivative test is used to find minimum and maximum values (extreme values) of a function of a single variable. If $f'\left( x \right) = 0$ at $x = a$ then $x = a$ is called a critical point. If $f''\left( x \right) > 0$ at critical point $x = a$ then we can say that $f\left( x \right)$ is minimum at $x = a$. If $f''\left( x \right) < 0$ at critical point $x = b$ then we can say that $f\left( x \right)$ is maximum at $x = b$.