Application of Derivatives 4 Question 31
33. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $8: 15$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are
(2013 Adv.)
(a) 24
(b) 32
(c) 45
(d) 60
$e^{x} \quad, \quad 0 \leq x \leq 1$
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Answer:
Correct Answer: 33. (b, c)
Solution:
- PLAN
The problem is based on the concept to maximise volume of cuboid, i.e. to form a function of volume, say $f(x)$ find $f^{\prime}(x)$ and $f^{\prime \prime}(x)$. Put $f^{\prime}(x)=0$ and check $f^{\prime \prime}(x)$ to be + ve or -ve for minimum and maximum, respectively
Here, $\quad l=15 x-2 a, b=8 x-2 a$ and $h=a$
$\therefore \quad$ Volume $=(8 x-2 a)(15 x-2 a) a$
$$ V=2 a \cdot(4 x-a)(15 x-2 a) $$
On differntiating Eq. (i) w.r.t $a$, we get
$$ \frac{d v}{d a}=6 a^{2}-46 a x+60 x^{2} $$
Again, differentiating,
$\frac{d^{2} v}{d a^{2}}=12 a-46 x$
| {f170454459647787899}$e^{x}$, if $0 \leq x \leq 1$ $2-e^{x-1}$, if $1<x \leq 2$ $x-e$, if $2<x \leq 3$ |
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|---|---|
| $g(x)=\int _0^{x} f(t) d t$ | |
| $g^{\prime}(x)=f(x)$ | |
| $g^{\prime}(x)=0 \Rightarrow x=1+\log _e 2$ and $x=e$. $e^{x}, \quad$ if $0 \leq x \leq 1$ |
|
 |
|
| $x=1+\log _e 2$ |
Here,
$$ \frac{d v}{d a}=0 \Rightarrow 6 x^{2}-23 x+15=0 $$
At $\quad a=5 \quad \Rightarrow \quad x=3, \frac{5}{6}$
$\Rightarrow \quad \frac{d^{2} v}{d a^{2}}=2(30-23 x)$
At $x=3, \frac{d^{2} v}{d a^{2}}=2(30-69)<0$
$\therefore \quad$ Maximum when $x=3$, also at $x=\frac{5}{6} \Rightarrow \frac{d^{2} v}{d a^{2}}>0$
$\therefore \quad$ At $x=5 / 6$, volume is minimum.
Thus, sides are $8 x=24$ and $15 x=45$
