Answer the question [ 4 mark question ]

Question 14 Marks

Find the maximum and minimum value of following function.$2 x^3-15 x^2+36 x+10$

Answer

Suppose $ y=2 x^3-15 x^2+36 x+10 $
So, $\frac{d y}{d x}=6 x^2-30 x+36$
and $\frac{d^2 y}{d x^2}=12 x-30$
For maxima and minima $\frac{d y}{d x}=0$
So, $6 x^2-30 x+36=0$
$\Rightarrow 6\left(x^2-5 x+6\right)=0$
$\Rightarrow x^2-5 x+6=0$
or $(x-3)(x-2)=0$
$\Rightarrow x=2,3$
now at $x=2, \frac{d^2 y}{d x^2}=12 \times 2-30=-6 < 0$
Function will maximum at $x=2$ and maximum value of function
$=2(2)^3-15(2)^2+36(2)+10$
$=16-60+72+10$
$=38$
again at $x=3, \frac{d^2 y}{d x^2}=12 \times 3-30=6>0$
Function will be minimum at $x=3$ and minimum value of function
$=2(3)^3-15(3)^2+36(3)+10$
$=2 \times 27-15 \times 9+108+10$
$=54-135+108+10$
$=37$

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Question 24 Marks

Surface area of a spherical bubble is increasing at the rate of $2 cm^2 / sec$. At what rate volume of bubble increasing when radius of bubble is $6 m$ ?

Answer

Suppose that radius of spherical bubble is $r$ and its surface area S and volume V .
then$
S=4 \pi r^2 \Rightarrow \frac{d S}{d r}=\frac{d}{d r}\left(4 \pi r^2\right)
$
$\Rightarrow \quad \frac{d S}{d r}=8 \pi r$
Rate of change of surface area w.r.t. time$
\frac{d S}{d t}=2 cm^2 / sec(\text { Given })
$
So, $\quad \frac{d S}{d t}=\frac{d S}{d r} \cdot \frac{d r}{d t}$
Put the values
$\Rightarrow \quad 2=8 \pi r \frac{d r}{d t}$
$\Rightarrow \quad \frac{d r}{d t}=\frac{2}{8 \pi r}=\frac{1}{4 \pi r}$
now, $\quad V =\frac{4}{3} \pi r^3 \Rightarrow \frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3} \pi r^3\right)$
$\Rightarrow \quad \frac{d V}{d r}=\frac{4}{3} \pi \cdot 3 r^2=4 \pi r^2$
So, $\quad \frac{d V}{d t}=\frac{d V}{d r} \times \frac{d r}{d t}$
Put the values
$
\frac{d V}{d t}=4 \pi r^2 \times \frac{1}{4 \pi r}=r
$
When $r=6 cm \frac{d V}{d t}=6 cm^3 / sec$

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