Question 11 Mark
Assertion $(A)$ : If two positive numbers are such that sum is $16$ and sum of their cubes is minimum, then numbers are $8, 8$.
Reason $(R)$ : If f be a function defined on an interval $I $ and $c \in l$ and let $f$ be twice differentiable at $c,$ then $x = c$ is a point of local minima if $f'(c) = 0$ and $f"(c) > 0$ and $f(c)$ is local minimum value of $f$.
Reason $(R)$ : If f be a function defined on an interval $I $ and $c \in l$ and let $f$ be twice differentiable at $c,$ then $x = c$ is a point of local minima if $f'(c) = 0$ and $f"(c) > 0$ and $f(c)$ is local minimum value of $f$.
Answer
View full question & answer→$(a)$ Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
Explanation : Let one number be $x,$ then the other number will be $(16-x).$
Let the sum of the cubes of these numbers be denoted by $S$.
Then, $S=x^3+(16-x)^3$
On differentiating $\text{w.r.t. x,}$ we get
$\frac{d S}{d z}=3 x ^2+3(16- x )^2(-1)$
$=3 x^2-3(16- x )^2$
$\Rightarrow \frac{d^2 S}{d x^2}=6 x+6(16- x )=96$
For minima put $\frac{d S}{d x}=0$.
$\therefore 3 x^2-3(16-x)^2=0$
$\Rightarrow x^2-\left(256+x^2-32 x\right)=0$
$\Rightarrow 32 x=256$
$\Rightarrow x=8$
At $x =8,\left(\frac{d^2 S}{d x^2}\right)_{x=8}=96 ; 0$
By second derivative test $, x = 8$ is the point of local minima of $S$.
Thus, the sum of the cubes of the numbers is the minimum when the numbers are $8$ and $16-8=8$
Hence, the required numbers are $8$ and $8.$
Explanation : Let one number be $x,$ then the other number will be $(16-x).$
Let the sum of the cubes of these numbers be denoted by $S$.
Then, $S=x^3+(16-x)^3$
On differentiating $\text{w.r.t. x,}$ we get
$\frac{d S}{d z}=3 x ^2+3(16- x )^2(-1)$
$=3 x^2-3(16- x )^2$
$\Rightarrow \frac{d^2 S}{d x^2}=6 x+6(16- x )=96$
For minima put $\frac{d S}{d x}=0$.
$\therefore 3 x^2-3(16-x)^2=0$
$\Rightarrow x^2-\left(256+x^2-32 x\right)=0$
$\Rightarrow 32 x=256$
$\Rightarrow x=8$
At $x =8,\left(\frac{d^2 S}{d x^2}\right)_{x=8}=96 ; 0$
By second derivative test $, x = 8$ is the point of local minima of $S$.
Thus, the sum of the cubes of the numbers is the minimum when the numbers are $8$ and $16-8=8$
Hence, the required numbers are $8$ and $8.$