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