Application of Derivatives: Maxima and Minima

A figure consists of a semicircle with a rectangle on its diameter. Given that the perimeter of the figure is $20 \; m$, find its dimensions in order that its area may be maximum.

Let $r$ be the radius of the semicircle.

Let the dimensions of the rectangle be $b$ and $2r$.

Perimeter of the figure $= \pi r + 2b + 2r$

Given: $\;$ Perimeter $= 20 \; m$

i.e. $\;$ $\pi r + 2b + 2r = 20$

i.e. $\;$ $r \left(2 + \pi\right) + 2b = 20$

i.e. $\;$ $b = 10 - \dfrac{r \left(2 + \pi\right)}{2}$ $\;\;\; \cdots \; (1)$

Area of the figure $= A = \dfrac{\pi r^2}{2} + 2 r b$

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 2 r \left[10 - \dfrac{r \left(2 + \pi\right)}{2}\right]$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 20 r - r^2 \left(2 + \pi\right)$ $\;\;\; \cdots \; (2)$

For maximum area, $\;$ $\dfrac{dA}{dr} = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Differentiating equation $(2)$ w.r.t $x$ we have,

$\dfrac{dA}{dr} = \dfrac{2 \pi r}{2} + 20 - 2r \left(2 + \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = \pi r + 20 - r \left(4 + 2 \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = 20 - r \left(\pi + 4\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ $\dfrac{dA}{dr} = 0$ $\implies$ $20 - r \left(\pi + 4\right) = 0$

i.e. $\;$ $r = \dfrac{20}{\pi + 4}$

From equation $(4)$, $\;$ $\dfrac{d^2 A}{dr^2} \Big |_{r = \frac{20}{\pi + 4}} = - \left(\pi + 4\right) < 0$

$\implies$ $r = \dfrac{20}{\pi + 4}$ $\;$ gives maximum area.

Substituting the value of $r$ in equation $(1)$ gives

$b = 10 - \dfrac{20 \left(\pi + 2\right)}{2 \left(\pi + 4\right)}$

i.e. $\;$ $b = 10 \left(1 - \dfrac{\pi + 2}{\pi + 4}\right) = \dfrac{20}{\pi + 4}$

$\therefore \;$ For maximum area of the figure, $\;$ radius of semicircle $= \dfrac{20}{\pi + 4} \; m$ $\;$ and the sides of the rectangle are $\;$ $\dfrac{20}{\pi + 4} \; m$, $\;$ $\dfrac{40}{\pi + 4} \; m$

Application of Derivatives: Maxima and Minima

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Let the two positive numbers be x and y.

Given: $x+y=16$

$\implies$ $y = 16-x$ $\;\; \cdots$ (1)

Sum of cubes of the numbers $= S = x^3 + y^3$

i.e. $S = x^3 + \left(16-x\right)^3$ [From equation (1)]

For minimum sum, $\dfrac{dS}{dx}=0$

Now, $\dfrac{dS}{dx}= 3x^2 - 3 \left(16-x\right)^2$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dS}{dx}=0$ $\implies$ $3x^2 - 3 \left(16-x\right)^2 = 0$

i.e. $x^2 = \left(16-x\right)^2$

i.e. $x^2 = 256 -32x + x^2$

i.e. $32x = 256$ $\implies$ $x = 8$

From equation (2), $\dfrac{d^2 S}{dx^2} = 6x +6 \left(16-x\right) = 96 > 0 \; \forall \; x$

$\implies$ $x = 8$ gives a minimum for the sum of cubes of the numbers.

From equation (1), when $x = 8$, $y=16-8=8$

$\therefore$ The two numbers are 8, 8.

Application of Derivatives: Maxima and Minima

What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere?

Let radius of sphere $=R$

Let radius of cylinder $=r$

Let height of cylinder $=2h$

From the figure, $R^2 = h^2 + r^2$

$\therefore$ $r^2 = R^2 - h^2$ $\;\; \cdots$ (1)

Volume of sphere $=V_s = \dfrac{4}{3}\pi R^3$ $\;\; \cdots$ (2)

Volume of cylinder $= V_c = 2\pi r^2 h$ $\;\; \cdots$ (3)

In view of equation (1), equation (3) becomes

$V_c = 2 \pi h \left(R^2 - h^2\right)$

i.e. $V_c = 2\pi R^2 h - 2 \pi h^3$ $\;\; \cdots$ (4)

For cylinder with maximum volume, $\dfrac{dV_c}{dh} = 0$

Now, from equation (4), $\dfrac{dV_c}{dh} = 2 \pi R^2 - 6\pi h^2$ $\;\; \cdots$ (5)

$\therefore$ $\dfrac{dV_c}{dh} = 0$ $\implies$ $2\pi R^2 - 6\pi h^2 = 0$

i.e. $3h^2 = R^2$ $\implies$ $h = \dfrac{R}{\sqrt{3}}$ $\;\; \cdots$ (6)

Now, from equation (5), $\dfrac{d^2 V_c}{dh^2} = -12 \pi h$

$\therefore$ $\dfrac{d^2 V_c}{dh^2} \bigg |_{h=\frac{R}{\sqrt{3}}} = \dfrac{-12\pi R}{\sqrt{3}} < 0$

$\implies$ $h = \dfrac{R}{\sqrt{3}}$ gives a cylinder with maximum volume.

Now, from equation (1), when $h = \dfrac{R}{\sqrt{3}}$,

$r = \sqrt{R^2 - \dfrac{R^2}{3}} = \dfrac{R\sqrt{2}}{\sqrt{3}}$ $\;\; \cdots$ (7)

$\therefore$ From equation (2), volume of cylinder $=V_c = 2 \pi \times \dfrac{2R^2}{3} \times \dfrac{R}{\sqrt{3}}$

i.e. $V_c = \dfrac{4\pi R^3}{3\sqrt{3}}$ $\;\; \cdots$ (8)

$\therefore$ From equations (2) and (8),

$\dfrac{V_c}{V_s} = \dfrac{4\pi R^3 / 3\sqrt{3}}{4\pi R^3 / 3} = \dfrac{1}{\sqrt{3}}$

$\therefore$ Fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere $= \dfrac{1}{\sqrt{3}} \times 100 = 57.5 \%$

Application of Derivatives: Maxima and Minima

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

Let the length of one piece of wire $= x$ cm

Then, length of remaining wire $= 36 - x$ cm

Let the wire of length x be turned in the form of a square.

$\therefore$ Length of side of square $= \dfrac{x}{4}$ cm

$\therefore$ Area of square $= \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16} \; cm^2$

$\left(36-x\right)$ cm wire is shaped as an equilateral triangle.

$\therefore$ Length of each side of triangle $=\dfrac{36-x}{3}$ cm

$\therefore$ Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \left(\dfrac{36-x}{3}\right)^2 = \dfrac{\sqrt{3}}{36} \left(36-x\right)^2 \; cm^2$

Sum of areas $= A = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}{36} \left(36-x\right)^2$

For minimum area, $\dfrac{dA}{dx}=0$

Now, $\dfrac{dA}{dx} = \dfrac{2x}{16} + \dfrac{\sqrt{3}}{36} \times 2 \times \left(36-x\right) \times \left(-1\right)$

i.e. $\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{\sqrt{3}}{18} \left(36-x\right)$ $\;\; \cdots$ (1)

$\therefore$ $\dfrac{dA}{dx} = 0$ $\implies$ $\dfrac{x}{8}- \dfrac{\sqrt{3}}{18} \left(36-x\right) = 0$

i.e. $9x = 4\sqrt{3}\left(36-x\right)$ $\implies$ $x = \dfrac{144 \sqrt{3}}{9+4\sqrt{3}}$

From equation (1), $\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{\sqrt{3}}{18} > 0$

$\implies$ $x = \dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ gives minimum area.

Length of second piece of wire $= 36 - \dfrac{144\sqrt{3}}{9+4\sqrt{3}} = \dfrac{324}{9+4\sqrt{3}}$

$\therefore$ Length of pieces of wire are $\dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ cm and $\dfrac{324}{9+4\sqrt{3}}$ cm

Application of Derivatives: Maxima and Minima

Find the area of the largest rectangle that fits inside a semicircle of radius 10 units. One side of the rectangle is along the diameter of the semicircle.

Let radius of semicircle $= r$

Let length of rectangle $= 2\ell$

Let breadth of rectangle $= b$

From the figure, $r^2 = \ell^2 + b^2$

$\implies$ $\ell = \sqrt{r^2 - b^2}$ $\;\; \cdots$ (1)

Area of rectangle $=A= 2\ell b$

Substituting the value of $\ell$ from equation (1) gives

$A = 2b \sqrt{r^2 -b^2}$

For maximum area, $\dfrac{dA}{db}=0$

Now, $\dfrac{dA}{db} = 2\sqrt{r^2 - b^2} + \dfrac{2b}{2 \sqrt{r^2-b^2}}\times \left(-2b\right)$

i.e. $\dfrac{dA}{db} = 2 \sqrt{r^2 - b^2} - \dfrac{2b^2}{\sqrt{r^2 - b^2}}$

i.e. $\dfrac{dA}{db} = \dfrac{2r^2 - 4b^2}{\sqrt{r^2 - b^2}}$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dA}{db} = 0$ $\implies$ $\dfrac{2r^2 -4b^2}{\sqrt{r^2 - b^2}} = 0$

i.e. $2b^2 = r^2$

$\implies$ $b = \dfrac{r}{\sqrt{2}}$

Substituting the value of b in equation (1) gives

$\ell = \sqrt{r^2 - \dfrac{r^2}{2}} = \dfrac{r}{\sqrt{2}}$

Now from equation (2) we have

$\dfrac{d^2 A}{db^2} = \dfrac{\sqrt{r^2 - b^2}\left(-2b\right)-\left(2r^2-4b^2\right) \times \dfrac{\left(-2b\right)}{2\sqrt{r^2-b^2}}}{r^2 - b^2}$

i.e. $\dfrac{d^2A}{db^2} = \dfrac{2b\left(r^2-2b^2-r^2+b^2\right)}{\left(r^2-b^2\right)\sqrt{r^2-b^2}}$

i.e. $\dfrac{d^2 A}{db^2} = \dfrac{-2b^3}{\left(r^2 - b^2\right) \sqrt{r^2 - b^2}}$

$\therefore$ $\dfrac{d^2A}{db^2} \bigg |_{b=\frac{r}{\sqrt{2}}} = \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\left(r^2 - \dfrac{r^2}{2}\right)\sqrt{r^2 - \dfrac{r^2}{2}}} $ $= \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\dfrac{r^3}{2\sqrt{2}}} = -2 < 0$

$\implies$ Area A is maximum when $\ell = \dfrac{r}{\sqrt{2}}$ and $b = \dfrac{r}{\sqrt{2}}$

$\therefore$ Maximum area of the rectangle $=2 \times \dfrac{r}{\sqrt{2}} \times \dfrac{r}{\sqrt{2}} = r^2$

Given: radius of semicircle $= r = 10$ units

$\therefore$ Maximum area of the rectangle $= 100$ sq units

Application of Derivatives: Maxima and Minima

Find the points at which the function f given by $f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$ has local maxima, local minima and point of inflection.

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

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

For critical points, $f'\left(x\right) = 0$

i.e. $\left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) = 0$

$\implies$ $x-2=0$ or $x+1=0$ or $7x-2=0$

i.e. $x=2$ or $x=-1$ or $x = \dfrac{2}{7}$

Consider $x=2$

When $x<2$, let $x=1.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(1.9\right) & = \left(1.9-2\right)^3 \left(1.9+1\right)^2 \left[\left(7\times 1.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

When $x>2$, let $x=2.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(2.1\right) & = \left(2.1-2\right)^3 \left(2.1+1\right)^2 \left[\left(7\times 2.1\right) -2\right] \\ & = \left(+ve\right) \left(+ve\right) \left(+ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from negative to positive as x increases through 2.

$\implies$ $x=2$ is a point of local minimum.

Consider $x=-1$

When $x<-1$, let $x=-1.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-1.1\right) & = \left(-1.1-2\right)^3 \left(-1.1+1\right)^2 \left[7 \times\left(-1.1\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>-1$, let $x=-0.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-0.9\right) & = \left(-0.9-2\right)^3 \left(-0.9+1\right)^2 \left[7 \times \left(-0.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ does not change sign as x increases through $-1$.

$\implies$ $x=-1$ is a point of inflexion.

Consider $x = \dfrac{2}{7}$

When $x < \dfrac{2}{7}$, let $x = \dfrac{1.9}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{1.9}{7}\right) & = \left(\dfrac{1.9}{7}-2\right)^3 \left(\dfrac{1.9}{7}+1\right)^2 \left[\left(7\times \dfrac{1.9}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>\dfrac{2}{7}$, let $x=\dfrac{2.1}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{2.1}{7}\right) & = \left(\dfrac{2.1}{7}-2\right)^3 \left(\dfrac{2.1}{7}+1\right)^2 \left[\left(7\times \dfrac{2.1}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from positive to negative as x increases through $\dfrac{2}{7}$.

$\implies$ $x=\dfrac{2}{7}$ is a point of local maximum.

Application of Derivatives: Maxima and Minima

A window is in the form of a rectangle surmounted by semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window for maximum air flow through the window.

Let width of window $= \ell$

Then diameter of semicircle $= \ell$

Let height of window $= b$

Perimeter of the window $= P = \ell + 2b + \dfrac{\pi \ell}{2}$

Given: Perimeter of window $= 10 \; m$

$\implies$ $\ell + 2b + \dfrac{\pi \ell}{2} = 10$

i.e. $\ell \left(2+\pi\right) + 4b = 20$

i.e. $b = 5 - \dfrac{\ell \left(2+\pi\right)}{4}$ $\;\; \cdots$ (1)

Area of the window $=A= b \; \ell + \dfrac{\pi}{2} \times \dfrac{\ell^2}{4}$ $\;\; \cdots$ (2)

Substituting the value of b from equation (1) in equation (2) gives

$A = \ell \left[5- \dfrac{\ell \left(2+\pi\right)}{4}\right] + \dfrac{\pi \ell^2}{8}$

i.e. $A= 5 \ell - \dfrac{\pi}{4} \ell^2 -\dfrac{1}{2}\ell^2 + \dfrac{\pi}{8} \ell^2$

i.e. $A = 5 \ell - \dfrac{\pi}{8}\ell^2 - \dfrac{1}{2}\ell^2$

i.e. $A = 5 \ell - \dfrac{\ell^2}{2}\left(\dfrac{\pi}{4}+1\right)$

For maximum air flow, $\dfrac{dA}{d\ell} = 0$

i.e. $5-\ell\left(\dfrac{\pi+4}{4}\right) = 0$

$\implies$ $\ell = \dfrac{20}{\pi+4}$

Substituting the value of $\ell$ in equation (1) gives

$b = 5 - \left(\dfrac{\pi + 2}{4}\right) \left(\dfrac{20}{\pi + 4}\right)$

i.e. $b = 5 - 5 \left(\dfrac{\pi + 2}{\pi + 4}\right)$

i.e. $b = \dfrac{5\pi + 20 - 5\pi -10}{\pi + 4}$

i.e. $b = \dfrac{10}{\pi + 4}$

$\therefore$ Length of the window $\ell = \dfrac{20}{\pi + 4} \; m$; breadth of the window $=b=\dfrac{10}{\pi + 4} \; m$

Application of Derivatives: Maxima and Minima

The total cost of manufacturing x pocket radios per day is ₹ $\left(\dfrac{x^2}{4}+35x+25\right)$ and rate at which they may be sold to a distributor is ₹ $\left(\dfrac{100-x}{2}\right)$ each. What should be the daily output to attain a maximum total profit.

Daily output $=x$ radios

Cost price of x radios $=CP= ₹ \left(\dfrac{x^2}{4}+35x+25\right) $

Sale price of x radios $= SP= ₹ \; x \left(50 - \dfrac{x}{2}\right) = ₹ \left(50x - \dfrac{x^2}{2}\right)$

$\therefore$ Profit function $P\left(x\right) = SP - CP= \left(50x - \dfrac{x^2}{2} - \dfrac{x^2}{4} - 35x - 25\right)$

i.e. $P\left(x\right) = 15x - \dfrac{3x^2}{4}-25$

$\therefore$ For maximum profit, $\dfrac{dP}{dx} = 0$

Now, $\dfrac{dP}{dx} = 15 - \dfrac{6x}{4}$

$\therefore$ $\dfrac{dP}{dx}=0$ $\implies$ $15 - \dfrac{3}{2}x = 0$

i.e. $\dfrac{3}{2}x = 15$ $\implies$ $x = 10$

$\therefore$ Daily output $=10$ radios.

Application of Derivatives: Maxima and Minima

$f\left(x\right) = x^3 + 3ax^2 +3bx + c$ has a maximum at $x=-1$ and a minimum zero at $x=1$. Find a, b and c.

$f\left(x\right) = x^3 + 3ax^2 + 3bx +c$

$f'\left(x\right) = 3x^2 + 6ax + 3b$

For maxima or minima, $f'\left(x\right) = 0$

$\implies$ $3x^2 + 6ax+3b = 0$

i.e. $x^2 + 2ax + b = 0$ $\;\; \cdots$ (1)

Now, $f\left(x\right)$ has a minimum zero at $x=1$

$\implies$ $f\left(1\right) = 0$

i.e. $1+3a+3b+c = 0$

i.e. $3a+3b+c = -1$ $\;\; \cdots$ (2)

Further, $f\left(x\right)$ has a maximum at $x=-1$ and minimum at $x=1$

$\implies$ $x = \pm 1$ are roots of quadratic equation (1)

$\therefore$ From equation (1), sum of roots $=\dfrac{2a}{1} = 1-1$

i.e. $2a = 0$ $\implies$ $a=0$

Also, from equation (1), product of roots $= \dfrac{b}{1} = 1 \times \left(-1\right)$

i.e. $b = -1$

Substituting the values of a and b in equation (2) give

$-3+c=-1$ $\implies$ $c=2$

$\therefore$ $a=0$, $b=-1$, $c=2$

Application of Derivatives: Maxima and Minima

A container holding a fixed volume is made in the shape of a cylinder with a hemispherical top. The hemispherical top has the same radius as the cylinder. Find the ratio of height to radius of the cylinder which minimizes the cost of the container if

1. the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with no bottom;


2. the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with a circular bottom, for which the cost per unit area is 1.5 times the cost per unit area of the side.

Let radius of cylinder = radius of hemisphere = r

Let height of cylinder = h

Volume of container $= V = \pi r^2 h + \dfrac{2}{3}\pi r^3$

i.e. $V = \pi r^2 \left(h+\dfrac{2}{3}r\right)$

i.e. $h+\dfrac{2}{3}r = \dfrac{V}{\pi r^2}$

$\implies$ $h = \dfrac{V}{\pi r^2} - \dfrac{2}{3}r$ $\;\; \cdots$ (1)

1. Let cost per unit area of the side $= c$
   
   Then, cost per unit area of the hemispherical top $=2c$
   
   Area of the side of the container $=2\pi r h$
   
   Area of the top of the container $= 2 \pi r^2$
   
   $\therefore$ Cost of the container $= C = 2\pi r h c + 4 \pi r^2 c$ $\;\; \cdots$ (2)
   
   Substituting the value of h from equation (1) in equation (2) gives
   
   $C = 2 \pi r c \left(\dfrac{V}{\pi r^2} - \dfrac{2}{3}r\right) + 4 \pi r^2 c$
   
   i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + 4\pi r^2 c$
   
   i.e. $C = 2cV \times \dfrac{1}{r} + \dfrac{8\pi c}{3} \times r^2$ $\;\; \cdots$ (3)
   
   For minimum cost, $\dfrac{dC}{dr}=0$
   
   $\therefore$ From equation (3), $\dfrac{dC}{dr} = \dfrac{-2cV}{r^2}+\dfrac{16 \pi c r}{3} = 0$
   
   i.e. $\dfrac{16 \pi c r}{3} = \dfrac{2cV}{r^2}$
   
   i.e. $r^3 = \dfrac{3V}{8\pi}$
   
   $\implies$ $\dfrac{V}{\pi r^3} = \dfrac{8}{3}$ $\;\; \cdots$ (4)
   
   Now, from equation (1),
   
   $\dfrac{h}{r} = \dfrac{V}{\pi r^3}- \dfrac{2}{3}$ $\;\; \cdots$ (5)
   
   In view of equation (4), equation (5) can be written as
   
   $\dfrac{h}{r}= \dfrac{8}{3}-\dfrac{2}{3} = 2$
   
   $\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=2:1$
   
   
2. Cost per unit area of the circular bottom of the container $= 1.5c$
   
   Area of the bottom of the container $= \pi r^2$
   
   $\therefore$ Cost of the container $=C = 4 \pi r^2 c + 2 \pi r h c + 1.5 \pi r^2 c$
   
   i.e. $C = 2\pi rhc + 5.5 \pi r^2 c$ $\;\; \cdots$ (6)
   
   Substituting the value of h from equation (1) in equation (6) gives
   
   $C= 2 \pi r c \left(\dfrac{V}{\pi r^2}-\dfrac{2}{3}r\right) + 5.5 \pi r^2 c$
   
   i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + \dfrac{11}{2}\pi r^2 c$
   
   i.e. $C = 2cV \times\dfrac{1}{r} + \dfrac{25\pi c}{6} \times r^2$ $\;\; \cdots$ (7)
   
   For minimum cost $\dfrac{dC}{dr} = 0$
   
   $\therefore$ We have from equation (7) for minimum cost
   
   $\dfrac{dC}{dr} = \dfrac{-2cV}{r^2} + \dfrac{50 \pi c r}{6} = 0$
   
   i.e. $\dfrac{2cV}{r^2} = \dfrac{25 \pi c r}{3}$
   
   $\implies$ $\dfrac{V}{\pi r^3} = \dfrac{25}{6}$ $\;\; \cdots$ (8)
   
   Substituting equation (8) in equation (5) gives
   
   $\dfrac{h}{r} = \dfrac{25}{6}- \dfrac{2}{3} = \dfrac{7}{2}$
   
   $\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=7:2$

Application of Derivatives: Maxima and Minima

Find the maximum and the minimum values of the function $f\left(x\right) = \sin x + \cos x$, $x \in \left[0,2\pi\right]$

$f\left(x\right) = \sin x + \cos x$

$\therefore$ $f'\left(x\right) = \cos x - \sin x$

For maxima or minima, $f'\left(x\right) = 0$

i.e. $\cos x - \sin x = 0$

i.e. $\sin x = \cos x$

i.e. $\tan x =1$

i.e. $x = \dfrac{\pi}{4} + k \pi$ where $k \in Z$

Since $x \in \left[0,2\pi\right]$, $f'\left(x\right) = 0$ $\implies$ $x = \dfrac{\pi}{4}, \; \dfrac{5\pi}{4}$

Now, $f''\left(x\right) = -\sin x - \cos x$

$\therefore$ $f''\left(\dfrac{\pi}{4}\right) = - \sin \left(\dfrac{\pi}{4}\right) - \cos \left(\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}} = -\sqrt{2} < 0$

$\implies$ $f\left(x\right)$ has local maximum at $x=\dfrac{\pi}{4}$ and $f\left(\dfrac{\pi}{4}\right)= \sqrt{2}$

$f''\left(\dfrac{5\pi}{4}\right) = - \sin \left(\dfrac{5\pi}{4}\right) - \cos \left(\dfrac{5\pi}{4}\right) = \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} = \sqrt{2} > 0$

$\implies$ $f\left(x\right)$ has a local minimum at $x = \dfrac{5\pi}{4}$ and $f\left(\dfrac{5\pi}{4}\right) = -\sqrt{2}$

For global maximum and minimum values consider $f\left(0\right)$ and $f\left(2\pi\right)$

Now, $f\left(0\right) = \sin 0 + \cos 0 = 1$

$f\left(2\pi\right) = \sin \left(2\pi\right) + \cos \left(2\pi\right) = 1$

Now $f\left(\dfrac{\pi}{4}\right) = \sqrt{2}$, $f\left(\dfrac{\pi}{2}\right) = 1$, $f\left(\dfrac{3\pi}{4}\right) = 0$, $f\left(\pi\right) = -1$, $f\left(\dfrac{5\pi}{4}\right) = -\sqrt{2}$, $f\left(\dfrac{3\pi}{2}\right) = -1$, $f\left(\dfrac{7\pi}{4}\right) = 0$

$\therefore$ Global maximum occurs at $x = \dfrac{\pi}{4}$ and global maximum $= f\left(\dfrac{\pi}{4}\right) = \sqrt{2}$

Global minimum occurs at $x = \dfrac{5\pi}{4}$ and global minimum $= f \left(\dfrac{5\pi}{4}\right) = -\sqrt{2}$

Application of Derivatives: Maxima and Minima

Find the dimensions of the rectangle of largest area having fixed perimeter 100.

Let length of rectangle $= \ell$

Let breadth of rectangle $= b$

Then, perimeter of rectangle $=2\left(\ell + b\right)$

Given: Perimeter of rectangle $= 100$

$\implies$ $2\left(\ell + b\right) = 100$

i.e. $\ell + b = 50$

$\implies$ $b = 50 - \ell$ $\;\; \cdots$ (1)

Area of rectangle $=A=\ell \; b$

i.e. $A = \ell \left(50 - \ell\right)$ $\; \;$ [in view of equation (1)]

i.e. $A = 50 \ell - \ell^2$ $\;\; \cdots$ (2)

We have from equation (2)

$\dfrac{dA}{d\ell} = 50 - 2\ell$ $\;\; \cdots$ (3)

For maximum area, $\dfrac{dA}{d\ell} = 0$

$\therefore$ From equation (3), $\dfrac{dA}{d\ell} = 50 - 2 \ell = 0$

$\implies$ $\ell = 25$

Now, from equation (3), $\dfrac{d^2 A}{d\ell^2} = -2 < 0$

$\implies$ $\ell = 25$ gives the length of a rectangle with maximum area.

Substituting the value of $\ell$ in equation (1) gives

$b= 50-25 = 25$

$\therefore$ Length of rectangle $= 25$ units; breadth of rectangle $= 25$ units

Application of Derivatives: Maxima and Minima

Find the maximum and the minimum values of the function $f\left(x\right) = 5-3x+5x^2 - x^3$, $x \in R$

$f\left(x\right) = 5-3x+5x^2-x^3$

$\therefore$ $f'\left(x\right) = -3 + 10x -3x^2$

For maxima or minima, $f'\left(x\right) = 0$

i.e. $-3+10x-3x^2 = 0$

i.e. $\left(3x-1\right)\left(x-3\right)=0$

$\therefore$ $f'\left(x\right) = 0$ $\implies$ $x=\dfrac{1}{3}$ or $x=3$

Now, $f''\left(x\right) = 10-9x$

$\therefore$ $f''\left(\dfrac{1}{3}\right) = 7 > 0$

$\implies$ f has local minimum at $x=\dfrac{1}{3}$ and maximum value $=f\left(\dfrac{1}{3}\right) = \dfrac{122}{27}$

$f''\left(3\right) = -17 < 0$

$\implies$ f has local maximum at $x=3$ and minimum value $= f\left(3\right) = 14$