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
$\frac{dP}{dy}=(35-y{)}^{2}5y^4+y^{5}2(35-y)(-1)$
$=y^4(35-y)[5(35-y)-2y]$
$=y^4(35-y)(175-5y-2y)$
$=y^4(35-y)(175-7y)=\left(35y^4-y^5\right)(175-7y)$
and $\frac{d^{2}P}{d{y}^2}=\left(35y^4-y^5\right)(-7)+(175-7y)\left(4\times 35\times y^3-5y^4\right)$
$=-7y^4(35-y)+7(25-y)\times 5y^3(28-y)$
$=-7y^4(35-y)+35y^3(25-y)(28-y)$
For maxima put $\frac{dP}{dy}=0$
$\Rightarrow y^4(35-y)(175-7y)=0$
$\Rightarrow y=0,35-y=0,175-7y=0$
$\Rightarrow y=0,y=25,y=35$
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.