Miscellaneous Exercise on Chapter 6

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Solutions

Question 1

Given:

$$ f(x)=\frac{\log x}{x} $$

Show that the function has a maximum at

$$ x=e. $$

Solution:

The function is

$$ f(x)=\frac{\log x}{x}. $$

Differentiate using the quotient rule.

Let

$$ u=\log x,\qquad v=x. $$

Then,

$$ u'=\frac1x,\qquad v'=1. $$

Hence,

$$ \begin{aligned} f'(x) &=\frac{vu'-uv'}{v^2}\\ &=\frac{x\left(\frac1x\right)-\log x}{x^2}\\ &=\frac{1-\log x}{x^2}. \end{aligned} $$

For critical points,

$$ f'(x)=0. $$

Therefore,

$$ 1-\log x=0. $$

$$ \log x=1. $$

Hence,

$$ \boxed{x=e.} $$

Now find the second derivative.

Using the quotient rule,

$$ \begin{aligned} f''(x) &=\frac{\left(-\frac1x\right)x^2-(1-\log x)(2x)} {x^4}\\ &=\frac{-x-2x(1-\log x)} {x^4}\\ &=\frac{2\log x-3}{x^3}. \end{aligned} $$

At

$$ x=e, $$

$$ \begin{aligned} f''(e) &=\frac{2(1)-3}{e^3}\\ &=-\frac1{e^3}<0. \end{aligned} $$

Therefore, the function has a maximum at

$$ \boxed{x=e.} $$

The maximum value is

$$ \begin{aligned} f(e) &=\frac{\log e}{e}\\ &=\frac1e. \end{aligned} $$

Answer:

• The function has a maximum at $$\boxed{x=e}$$.
• The maximum value is $$\boxed{\dfrac1e}$$.

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Question 2

Given:

The two equal sides of an isosceles triangle with fixed base

$$ b $$

are decreasing at the rate of

$$ 3\text{ cm/s}. $$

Find the rate at which the area is decreasing when the two equal sides are equal to the base.

Solution:

Let

• Equal side = $$x$$ cm
• Base = $$b$$ cm (constant)
• Height = $$h$$ cm
• Area = $$A$$

Since the altitude bisects the base, by Pythagoras theorem,

$$ \begin{aligned} h &=\sqrt{x^2-\left(\frac b2\right)^2}. \end{aligned} $$

Therefore, the area is

$$ \begin{aligned} A &=\frac12bh\\ &=\frac b2\sqrt{x^2-\frac{b^2}{4}}. \end{aligned} $$

Differentiate both sides with respect to time \(t\).

$$ \begin{aligned} \frac{dA}{dt} &=\frac b2\cdot \frac{1}{2\sqrt{x^2-\frac{b^2}{4}}} \cdot2x\frac{dx}{dt}. \end{aligned} $$

Hence,

$$ \boxed{ \frac{dA}{dt} = \frac{bx}{2\sqrt{x^2-\frac{b^2}{4}}} \cdot \frac{dx}{dt}. } $$

It is given that the equal sides are decreasing at the rate

$$ \frac{dx}{dt}=-3\text{ cm/s}. $$

When the equal sides are equal to the base,

$$ x=b. $$

Therefore,

$$ \begin{aligned} \frac{dA}{dt} &= \frac{b^2}{2\sqrt{b^2-\frac{b^2}{4}}} (-3). \end{aligned} $$

Simplify the denominator.

$$ \begin{aligned} \sqrt{b^2-\frac{b^2}{4}} &=\sqrt{\frac{3b^2}{4}}\\ &=\frac{\sqrt3\,b}{2}. \end{aligned} $$

Substituting,

$$ \begin{aligned} \frac{dA}{dt} &= \frac{b^2}{\sqrt3\,b} (-3)\\ &= -\sqrt3\,b. \end{aligned} $$

Thus, the area is decreasing at the rate

$$ \boxed{\sqrt3\,b\text{ cm}^2/\text{s}.} $$

Answer:

$$ \boxed{ \frac{dA}{dt} = -\sqrt3\,b\text{ cm}^2/\text{s} } $$

Hence, the area decreases at the rate

$$ \boxed{\sqrt3\,b\text{ cm}^2/\text{s}.} $$

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Question 5

Given:

Find the maximum area of an isosceles triangle inscribed in the ellipse

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$

with its vertex at one end of the major axis.

Solution:

Assume the major axis is along the \(x\)-axis and let the vertex of the triangle be

$$ A(a,0). $$

Let the other two vertices be

$$ B(x,y)\quad\text{and}\quad C(x,-y), $$

where \(B\) and \(C\) lie on the ellipse.

The base of the triangle is

$$ BC=2y. $$

The perpendicular distance from \(A(a,0)\) to the line \(BC\) is

$$ a-x. $$

Hence, the area of the triangle is

$$ \begin{aligned} A &=\frac12\times2y\times(a-x)\\ &=y(a-x). \end{aligned} $$

From the equation of the ellipse,

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, $$

we obtain

$$ y=b\sqrt{1-\frac{x^2}{a^2}}. $$

Therefore,

$$ A(x)=b(a-x)\sqrt{1-\frac{x^2}{a^2}}. $$

To simplify differentiation, maximize

$$ A^2. $$

$$ \begin{aligned} A^2 &=b^2(a-x)^2\left(1-\frac{x^2}{a^2}\right). \end{aligned} $$

Since

$$ 1-\frac{x^2}{a^2} = \frac{(a-x)(a+x)}{a^2}, $$

we get

$$ A^2 = \frac{b^2}{a^2}(a-x)^3(a+x). $$

Differentiate.

$$ \begin{aligned} \frac{d(A^2)}{dx} &=\frac{b^2}{a^2} \left[-3(a-x)^2(a+x)+(a-x)^3\right]. \end{aligned} $$

Factorising,

$$ \frac{d(A^2)}{dx} = -\frac{2b^2}{a^2} (a-x)^2(a+2x). $$

For maximum area,

$$ a+2x=0. $$

Hence,

$$ \boxed{x=-\frac a2.} $$

Now find the corresponding value of \(y\).

$$ \begin{aligned} y &= b\sqrt{1-\frac{1}{4}}\\ &=\frac{\sqrt3\,b}{2}. \end{aligned} $$

Therefore,

$$ \begin{aligned} A_{\max} &= \frac{\sqrt3\,b}{2} \left(a+\frac a2\right)\\ &= \frac{\sqrt3\,b}{2} \cdot \frac{3a}{2}\\ &= \boxed{\frac{3\sqrt3}{4}ab.} \end{aligned} $$

Answer:

$$ \boxed{ \text{Maximum Area} = \frac{3\sqrt3}{4}ab } $$

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Question 6

Given:

A tank with rectangular base and rectangular sides, open at the top, is to be constructed.

• Depth = $$2\text{ m}$$
• Volume = $$8\text{ m}^3$$
• Cost of base = Rs. $$70$$ per m²
• Cost of sides = Rs. $$45$$ per m²

Find the cost of the least expensive tank.

Solution:

Let

• Length = $$x$$ m
• Breadth = $$y$$ m
• Depth = $$2$$ m

The volume is

$$ 2xy=8. $$

Therefore,

$$ xy=4. $$

Hence,

$$ y=\frac4x. $$

Cost of the base

Area of the base is

$$ xy=4\text{ m}^2. $$

Therefore,

$$ \text{Base Cost} = 70\times4 = 280. $$

Cost of the four sides

Total area of the four sides is

$$ 2(x\times2)+2(y\times2) = 4(x+y). $$

Hence,

$$ \text{Side Cost} = 45\times4(x+y) = 180(x+y). $$

Substitute

$$ y=\frac4x. $$

$$ \begin{aligned} C(x) &=280+180\left(x+\frac4x\right). \end{aligned} $$

Differentiate.

$$ \begin{aligned} C'(x) &=180-\frac{720}{x^2}. \end{aligned} $$

For minimum cost,

$$ 180-\frac{720}{x^2}=0. $$

$$ 180x^2=720. $$

$$ x^2=4. $$

Since length is positive,

$$ x=2. $$

Then

$$ y=\frac42=2. $$

Now find the second derivative.

$$ C''(x)=\frac{1440}{x^3}. $$

At

$$ x=2, $$

$$ C''(2)=180>0. $$

Hence, the cost is minimum.

The minimum cost is

$$ \begin{aligned} C(2) &=280+180(2+2)\\ &=280+720\\ &=1000. \end{aligned} $$

Answer:

• Length = $$2\text{ m}$$
• Breadth = $$2\text{ m}$$
• Depth = $$2\text{ m}$$
• Minimum cost = Rs. 1000

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Question 7

Given:

The sum of the perimeter of a circle and a square is

$$ k, $$

where \(k\) is a constant.

Show that the sum of their areas is least when the side of the square is double the radius of the circle.

Solution:

Let

• Radius of the circle = $$r$$
• Side of the square = $$x$$

The given condition is

$$ 2\pi r+4x=k. $$

Hence,

$$ r=\frac{k-4x}{2\pi}. $$

The total area is

$$ A=\pi r^2+x^2. $$

Substituting the value of \(r\),

$$ \begin{aligned} A(x) &=\pi\left(\frac{k-4x}{2\pi}\right)^2+x^2\\ &=\frac{(k-4x)^2}{4\pi}+x^2. \end{aligned} $$

Differentiate with respect to \(x\).

$$ \begin{aligned} A'(x) &=\frac{2(k-4x)(-4)}{4\pi}+2x\\ &=-\frac{2(k-4x)}{\pi}+2x. \end{aligned} $$

For minimum area,

$$ A'(x)=0. $$

Therefore,

$$ -\frac{2(k-4x)}{\pi}+2x=0. $$

$$ \frac{k-4x}{\pi}=x. $$

Since

$$ k-4x=2\pi r, $$

we obtain

$$ 2\pi r=\pi x. $$

Hence,

$$ \boxed{x=2r.} $$

Now find the second derivative.

$$ \begin{aligned} A''(x) &=\frac{8}{\pi}+2. \end{aligned} $$

Since

$$ A''(x)>0, $$

the total area is minimum.

Answer:

$$ \boxed{\text{The total area is least when }x=2r.} $$

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Question 8

Given:

A window consists of a rectangle surmounted by a semicircle.

The total perimeter of the window is

$$ 10\text{ m}. $$

Find the dimensions of the window so that the area is maximum.

Solution:

Let

• Radius of the semicircle = $$r$$ m
• Height of the rectangular portion = $$h$$ m

The width of the rectangle is

$$ 2r. $$

The perimeter of the window is

$$ 2h+2r+\pi r=10. $$

Hence,

$$ h=\frac{10-(2+\pi)r}{2}. $$

The total area is

$$ A=2rh+\frac12\pi r^2. $$

Substituting the value of \(h\),

$$ \begin{aligned} A(r) &=2r\left(\frac{10-(2+\pi)r}{2}\right) +\frac12\pi r^2\\ &=10r-(2+\pi)r^2+\frac12\pi r^2\\ &=10r-\left(2+\frac{\pi}{2}\right)r^2. \end{aligned} $$

Differentiate.

$$ \begin{aligned} A'(r) &=10-(4+\pi)r. \end{aligned} $$

For maximum area,

$$ 10-(4+\pi)r=0. $$

Therefore,

$$ \boxed{ r=\frac{10}{4+\pi}. } $$

Now substitute this value into the expression for \(h\).

$$ \begin{aligned} h &=\frac{10-(2+\pi)\left(\frac{10}{4+\pi}\right)}{2}\\ &=\frac{10}{4+\pi}. \end{aligned} $$

Thus,

$$ h=r. $$

Now find the second derivative.

$$ A''(r)=-(4+\pi). $$

Since

$$ A''(r)<0, $$

the area is maximum.

Therefore,

• Radius of the semicircle = $$\displaystyle \frac{10}{4+\pi}\text{ m}$$
• Height of the rectangular part = $$\displaystyle \frac{10}{4+\pi}\text{ m}$$
• Width of the window = $$\displaystyle \frac{20}{4+\pi}\text{ m}$$

Answer:

$$ \boxed{ \begin{aligned} r&=\frac{10}{4+\pi}\text{ m},\\ h&=\frac{10}{4+\pi}\text{ m},\\ \text{Width}&=\frac{20}{4+\pi}\text{ m}. \end{aligned} } $$

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Question 9

Given:

A point on the hypotenuse of a right-angled triangle is at distances

$$ a $$

and

$$ b $$

from the two sides containing the right angle.

Show that the minimum length of the hypotenuse is

$$ \left(a^{\frac23}+b^{\frac23}\right)^{\frac32}. $$

Solution:

Let the lengths of the two perpendicular sides be

$$ x \quad \text{and} \quad y. $$

Then the hypotenuse is

$$ L=\sqrt{x^2+y^2}. $$

The equation of the hypotenuse joining

$$ (x,0) $$

and

$$ (0,y) $$

is

$$ \frac{X}{x}+\frac{Y}{y}=1. $$

Since the given point is at distances

$$ a $$

and

$$ b $$

from the two perpendicular sides, its coordinates are

$$ (a,b). $$

As the point lies on the hypotenuse,

$$ \frac{a}{x}+\frac{b}{y}=1. $$

Hence,

$$ y=\frac{bx}{x-a}. $$

The square of the hypotenuse is

$$ L^2=x^2+y^2. $$

Substituting the value of \(y\),

$$ L^2=x^2+\frac{b^2x^2}{(x-a)^2}. $$

Differentiating and simplifying,

$$ \frac{d(L^2)}{dx}=0 $$

gives

$$ \frac{x-a}{a} = \left(\frac{b}{a}\right)^{\frac23}. $$

Hence,

$$ x = a+a^{\frac13}b^{\frac23}. $$

Similarly,

$$ y = b+a^{\frac23}b^{\frac13}. $$

Therefore,

$$ \begin{aligned} L^2 &=x^2+y^2\\ &=\left(a^{\frac23}+b^{\frac23}\right)^3. \end{aligned} $$

Hence,

$$ \boxed{ L= \left(a^{\frac23}+b^{\frac23}\right)^{\frac32}. } $$

The second derivative is positive at the critical point, therefore the value obtained is the minimum length of the hypotenuse.

Hence proved.

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Question 10

Given:

$$ f(x)=(x-2)^4(x+1)^3 $$

Find:

• (i) Local maxima
• (ii) Local minima
• (iii) Point of inflexion

Solution:

Differentiate using the product rule.

$$ \begin{aligned} f'(x) &=4(x-2)^3(x+1)^3 +3(x-2)^4(x+1)^2. \end{aligned} $$

Taking common factors,

$$ \begin{aligned} f'(x) &=(x-2)^3(x+1)^2 \left[4(x+1)+3(x-2)\right]\\ &=(x-2)^3(x+1)^2(7x-2). \end{aligned} $$

The critical points are

$$ x=-1,\qquad x=\frac27,\qquad x=2. $$

Sign of $$f'(x)$$

Interval	Sign of $$f'(x)$$
$$(-\infty,-1)$$	Positive
$$(-1,\frac27)$$	Positive
$$\left(\frac27,2\right)$$	Negative
$$(2,\infty)$$	Positive