Calculus Solution Ex # 4.5

Exercise 4.5

Question 1
Let $x$ be the required number, then the reciprocal is $\frac{1}{x}$ .

$f(x)=x+\frac{1}{x};\quad\left[\frac{1}{2}, 3 / 2\right]--------(1)$

differentiating (1) w.r.t $x$

$f^{\prime}(x)=1-\frac{1}{x^2}$

Set

$f^{\prime}(x)=0$

$\Rightarrow 1-\frac{1}{x^2}=0 \Rightarrow 1=\frac{1}{x^2}$

$\Rightarrow x^2=1, \quad \Rightarrow x= \pm 1$

But only $x=1$ lie in the interval $\left(\frac{1}{2}, \frac{3}{2}\right)$ , therefore neglecting $x=-1$ .
Now substituting $x=\frac{1}{2}, 1$ , and $\frac{3}{2}$ in (1)

$f(1 / 2)=\frac{1}{2}+\frac{1}{1 / 2}=\frac{1}{2}+2=5 / 2$

$f(1)=1+\frac{1}{1}=2$

$f\left(\frac{3}{2}\right)=\frac{3}{2}+\frac{1}{3 / 2}=\frac{3}{2}+\frac{2}{3}=\frac{9+5}{6}=\frac{13}{6}$

(a) $f$ is as small as possible at x=1

(b) $f$ is as large as possible at $x=\frac{1}{2}.$

Question 2
Let $x$ and $y$ be the two required numbers then

$x+y=1-------(2)$

$f=x^2+y^2-----(2)$

From (1) we have $\quad y=1-x$
substituting $y=1-x$ into (2) to get $f$ as a function one singe variable $x$

$f=x^2+(1-x)^2$

$f=x^2+1+x^2-2 x$

$f=2 x^2-2 x+1 ; \quad 0 \leq x \leq 1$

Differentiating w.r.t. $x$

$f^{\prime}(x)=4 x-2----(3)$

Now we shall put $f^{\prime}(x)=0$ to find the values of $x$ which lies in the interval $[0,1]$ . Therefore, from (3) we have

$f^{\prime}(x)=0$

$\Rightarrow 4x-2=0$

$\Rightarrow 4x=2$

$\Rightarrow x=2/4$

$\Rightarrow x=1/2$

Thus $x=1/2$ lies on the interval $(0,1)$

We shall find the values of $f(x)$ at the end point of the interval $|0,1]$ ie $x=0, 1$ and $x= \frac{1}{2}$ .

$f(0)=0-0+1=1$

$f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^2-2\left(\frac{1}{2}\right)+1=\frac{1}{2}-1+1=\frac{1}{2}$

$f(1)=2(1)^2-2(1)+1=1$

a) $f$ is maximum at $x=0$ , and at $x=1$
so
when $x=0$ , then $y=1-0=1$
when $x=1$ , then $y=1-1=0$

Thus we conclude that $f$ is as large as possible when one of the number is $0$ and other is $1$ .

b) $f$ is minimum when $x=1/2$
So since $x=\frac{1}{2}$ then $y=1-\frac{1}{2}=\frac{1}{2}$ .

Question 3
Let $x$ be the length and $y$ breadth of the rectangular feiled then

$x+2y=1000$

$2y=1000-x$

$y=500-x / 2------(1)$

Let $A$ be we area of the rectangular field

$A=\text{length}\times \text{breadth}$

$A=xy-----(2)$

By using (1), we have

$A=x\left(500-\frac{x}{2}\right)$

$A=500x-\frac{x^2}{2} \quad\quad 0 \leq x \leq 100-----(3)$

differentiating w.r.t $x$

$\frac{d A}{d x}=500-{x}$

$\begin{aligned} \text {Set} \quad \quad \frac{d A}{d x}=0 \Rightarrow \quad 500 -{x}=0 \end{aligned}$

$\Rightarrow x=500$

substituting $x=0$ , 500 and 1000 into (3)

$A(0)=0$

$A(500)=500(500)-\frac{(500)^2}{2}=250000-\frac{250000}{2}$

$=\frac{250000}{2}=125000$

$A(1000)=500(1000)-\frac{(1000)^2}{2}$

$A(1000)=500000-500000=0$

Thus $A$ is maximum when $x=500$
Substituting x=500 in to (1) we get

$y=500-\frac{500}{2}=500-250=250$