In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.
Solved example 22.82
A circular disc of radius 3 cm is being heated. Due to expansion, it's radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.
Solution:
1. We have the formula for area of a circle: A = πr2.
2. The radius increases at the rate of 0.05 cm/s.
•
So radius at time t will be 0.05t
4. So area A at time t = π(0.05t)2 = 0.0025πt2 cm2
•
Now the rate of change of area w.r.t radius can be obtained as:
dA/dt = 0.0025π(2t) = 0.005 πt cm2/s
•
This result can be used to find the rate of change of area at any instant.
5. We want the rate when radius is 3.2 cm. So we want the instant at which the radius is 3.2 cm. For that, we can use the result in (2). We get:
3.2 = 0.05t
⇒ t = 64
6. So from (4) we get:
(dA/dt)t=64 = 0.005π(64) = 0.320π cm2/s
Solved example 22.83
An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and
folding up the sides. Find the volume of the largest such box.
Solution:
Step I: Writing the problem as a function
1. In fig.22.77 below, the original rectangular sheet is shown in yellow color. Red squares are cut off from the corners.
 |
| Fig.22.77 |
• So when the flaps are folded up, we will get a box with:
♦ base a rectangle (8 − 2x) × (3 − 2x) m2
♦ height x m
• Then the volume of the box = x(8 − 2x)(3 − 2x) m3
2. It is clear that, V depends on x. That means, V is a function of x. So we can write:
V = f(x) = x(8 − 2x)(3 − 2x) m3
3. We want V to be as large as possible. That means, we want the largest of all local maxima of f. That means, we need to find all local maxima and compare them.
Step II: Finding the critical points
1. Write the first two derivatives:
•
f '(x) =12x2 − 44x + 24
• f ''(x) = 24x − 44
2. Equate the first derivative to zero and solve for x:
12x2 − 44x + 24 = 0
⇒ x = 2/3, x = 3
3. So the only two points in category I are: x = 2/3 and x = 3
4. We obtained f '(x) = 12x2 − 44x + 24
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only two critical points are: x = 2/3 and x = 3
Step III: Find f ''(x) at the critical points
1. f ''(2/3) = 24(2/3) − 44 = −28
2. f ''(3) = 24(3) − 44 = 28
Step IV: Applying the second derivative test
1. Since f ''(2/3) is −ve, there is a local maximum at x = 2/3
2. Since f ''(3) is +ve, there is a local minimum at x = 3
Step V: Finding the actual maximum value:
• The local maximum value at x = 2/3 is given by:
f(2/3) = (2/3)[8 − 2(2/3)][3 − 2(2/3)] = 200/27 m3.
Solved example 22.84
Manufacturer can sell x items at a price of rupees $\rm{5 - \frac{x}{100}}$ each. The cost price of x items is Rs $\rm{\frac{x}{5} + 500}$. Find the number of items he should sell to earn maximum profit.
Solution:
Step I: Writing the problem as a function
1. When the manufacturer sells x items, he will get an income of: Rs $\rm{x\left(5 - \frac{x}{100} \right)}$
2. To produce those x items, he would have spent:
Rs $\rm{\frac{x}{5} + 500}$
3. So the profit P can be obtained as:
$\rm{P\,=\, x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
4. It is clear that, P depends on x. That means, P is a function of x. So we can write:
$\rm{P\,=\, f(x)\,=\,x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
5.
We want P to be as large as possible. That means, we want the largest
of all local maxima of f. That means, we need to find all local maxima
and compare them.
Step II: Finding the critical points
1. Write the first two derivatives:
•
f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$
• f ''(x) = $\rm{\frac{-1}{50}}$
2. Equate the first derivative to zero and solve for x:
$\rm{\frac{-x}{50} + \frac{25}{4} \,=\, 0}$
⇒ x = 240
3. So the only one point in category I is: x = 240
4. We obtained f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 240
Step III: Find f ''(x) at the critical points
• f ''(x) is a constant: $\rm{\frac{-1}{50}}$
Step IV: Applying the second derivative test
• Since f ''(240) is −ve, there is a local maximum at x = 240
• For maximum profit, the manufacturer must sell 240 items.
Solved example 22.85
Find the maximum area of an isosceles triangle inscribed in the ellipse $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$ with it's vertex at one end of the major axis.
Solution:
Step I: Writing the problem as a function
1. The given ellipse is: $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$
• So it is clear that:
♦ Major axis lies along the x-axis.
♦ Minor axis lies along the y-axis.
♦ Center of ellipse is at the origin O.
2. Thus we get fig.22.78 below:
 |
| Fig.22.78 |
• The vertex A of the isosceles triangle is at one end of the major axis.
3. Let the base of the isosceles triangle intersect the x-axis at D(h,0).
• Then the y-coordinates of B and C can be obtained as shown below:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}} & {~=~} &{1} \\
{~\color{magenta} 2 } &{\implies} &{\frac{h^2}{a^2}\,+\,\frac{y^2}{b^2}} & {~=~} &{1} \\
{~\color{magenta} 3 } &{\implies} &{\frac{y^2}{b^2}} & {~=~} &{1~-~\frac{h^2}{a^2}} \\
{~\color{magenta} 4 } &{\implies} &{y^2} & {~=~} &{b^2 \left(1~-~\frac{h^2}{a^2} \right)} \\
{~\color{magenta} 5 } &{\implies} &{y} & {~=~} &{\pm \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}} \\
\end{array}$
• So the y-coordinate of B is $\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• And the y-coordinate of C is $-\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• Then the distance BC will be $2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
4. Next we want the coordinates of A. It can be calculated as shown below:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}} & {~=~} &{1} \\
{~\color{magenta} 2 } &{\implies} &{\frac{x^2}{a^2}\,+\,\frac{0^2}{b^2}} & {~=~} &{1} \\
{~\color{magenta} 3 } &{\implies} &{\frac{x^2}{a^2}} & {~=~} &{1} \\
{~\color{magenta} 4 } &{\implies} &{x^2} & {~=~} &{a^2} \\
{~\color{magenta} 5 } &{\implies} &{x} & {~=~} &{\pm a} \\
\end{array}$
• So the coordinates of A are: (a,0).
• Then the altitude AD = (h+a)
5.
Now we can write the expression for the area of triangle ABC:
Area = 1/2 × Base × Altitude = 1/2 × BC × AD
= $\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)$
• This can be simplified as:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Area}} & {~=~} &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{b \sqrt{\left(1~-~\frac{h^2}{a^2} \right)} × (h+a)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{b \sqrt{\left(\frac{a^2 – h^2}{a^2} \right)} × (h+a)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)} \\
\end{array}$
6. It is clear that, Area A depends on h. That means, A is a function of h. So we can write:
$\rm{A\,=\, f(x)\,=\,\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}$
5.
We want A to be as large as possible. That means, we want the largest
of all local maxima of f. That means, we need to find all local maxima
and compare them.
Step II: Finding the critical points
1. Write the first two derivatives:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f(h)} & {~=~} &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)} \\
{~\color{magenta} 2 } &{\implies} &{f'(h)} & {~=~} &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)} × (1) ~+~(h+a) (1/2) \left(a^2 – h^2 \right)^{-1/2} (-2h) \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)} ~-~ \frac{h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – h^2 ~-~h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – h^2 – h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
\end{array}$
2. Equate the first derivative to zero and solve for h:
$\rm{a^2 – 2 h^2 – ha \,=\, 0}$
⇒ $\rm{2 h^2 + ha - a^2 \,=\, 0}$
• Using quadratic formula, we get:
h = a/2 and h = −a
• h cannot be −a. Because, then in fig.22.78 above, D will coincide with the left end of the major axis. Parts of the triangle will fall outside the ellipse.
3. So the only one point in category I is: x = a/2
4. We obtained f '(h) = $\frac{b}{a} \left[\frac{a^2 – 2 h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]$
• This function is not defined when h = a. But when h = a, the function itself becomes invalid because there will be no inscribed triangle.
• Therefore, there will be no points in category II.
5. So the only one critical point is: h = a/2
6. The domain is (−a,a).
• Based on the critical point, the domain can be divided into two intervals: (−a, a/2) and (a/2, a)
Step III: Analyzing the signs of f '(x) in the two intervals
1. Consider the first interval. Zero is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(h)} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
{~\color{magenta} 2 } &{\implies} &{f'(0)} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 (0)^2 – (0)a}{\sqrt{\left(a^2 – (0)^2 \right)}} \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{a^2}{\sqrt{\left(a^2 \right)}} \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{b} \\
\end{array}$
• So in the first interval. f '(x) is +ve.
2. Consider the second interval. $\frac{3a}{4}$ is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(h)} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]} \\
{~\color{magenta} 2 } &{\implies} &{f'(3a/4)} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 (3a/4)^2 – (3a/4)a}{\sqrt{\left(a^2 – (3a/4)^2 \right)}} \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{a^2 – 2 \left(\frac{9 a^2}{16} \right) – \frac{3 a^2}{4}}{\sqrt{\left(a^2 – \frac{9 a^2}{16} \right)}} \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 18 a^2 – 12 a^2}{16}}{\sqrt{\left(\frac{16 a^2 – 9 a^2}{16} \right)}} \right]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 30 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\frac{b}{a} \left[\frac{\frac{- 14 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]} \\
\end{array}$
• So in the second interval. f '(x) is −ve.
Step IV: Applying the first derivative test
• Since f '(x) changes sign from +ve to −ve at the critical point, we have a maximum at that critical point.
Step V: Final result:
• For maximum area, h = a/2
So from step I, we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Area}} & {~=~} &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)} \\
{~\color{magenta} 2 } &{\implies} &{\text{Area (max)}} & {~=~} &{\sqrt{b^2 \left(1~-~\frac{(a/2)^2}{a^2} \right)} × ((a/2)+a)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\sqrt{b^2 \left(1~-~\frac{a^2 / 4}{a^2} \right)} × (3a/2)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\sqrt{3 b^2 / 4} × (3a/2)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{\sqrt{3} b}{2} × \frac{3a}{2}} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\frac{3 \sqrt{3} \,ab}{4}} \\
\end{array}$
The link below gives a few more miscellaneous examples:
We have completed this chapter on applications of derivatives. In the next chapter, we will see Integrals.
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