Question

Statement I If two positive numbers $x$ and $y$ are such that $x+y=35$ and $x^2 y^5$ is maximum, then the numbers are 10 and 25 .
Statement II If $f$ be a function defined on an interval $I$ and $c \in I$ and also, if $f$ be twice differentiable at $c$, then $x=c$ is a point of local maxima if $f^{\prime}(c)=0$ and $f^{\prime \prime}(c)<0$ and the value $f(c)$ is local maximum value of $f$.

• A. Statement I is true, statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement I is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (A)

Let the numbers be $x$ and $y$ and $P=x^2 y^5$, then $ x+y=35 \Rightarrow x =35-y $
$ \therefore P =(35-y)^2 y^5$
On differentiating twice w.r.t. $y$, we get
$\frac{d P}{d y} =(35-y)^2 5 y^4+y^5 2(35-y)(-1) $
$=y^4(35-y)[5(35-y)-2 y]$
$ =y^4(35-y)(175-5 y-2 y) $
$ =y^4(35-y)(175-7 y)=\left(35 y^4-y^5\right)(175-7 y)$
and $\frac{d^2 P}{d y^2}=\left(35 y^4-y^5\right)(-7)+(175-7 y)\left(4 \times 35 \times y^3-5 y^4\right)$
$=-7 y^4(35-y)+7(25-y) \times 5 y^3(28-y)$
$=-7 y^4(35-y)+35 y^3(25-y)(28-y)$
For maxima put $\frac{d P}{d y}=0$
$\Rightarrow y^4(35-y)(175-7 y)=0 $
$\Rightarrow y=0,35-y=0,175-7 y=0$
$\Rightarrow y=0, y=25, y=35$
When $y=0, x=35-0=35$ and the product $x^2 y^5$ will be 0 .
When $y=35$ and $x=35-35=0$. This will make the product $x^2 y^5$ equal to 0 .
$\therefore y=0$ and $y=35$ cannot be the possible value of $y$.
When $y=25,\left(\frac{d^2 P}{d y^2}\right)_{y-25}=-7 \times(25)^4 \times(35-25)+35$
$ \times(25)^3 \times(25-25)(28-25) $
$=-7 \times 390625 \times 10+35 \times 15625 \times 0 \times 3$
$=-27343750+0=-27343750 < 0$
$\therefore$ By second derivative test, $P$ will be the maximum when $y=25$ and $x=35-25=10$.
Hence, the required numbers are 10 and 25.