Application of Derivative Exercise: 18.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 18.1

1. (i) Examine whether the function f(x) = 15x² – 14x + 1 is increasing or decreasing at x=$\frac{2}{5}$ and x=$\frac{5}{2}$.
Solution:

Given f(x) = 15x² – 14x + 1

f'(x) = 30x – 14

At x = $\frac{2}{5}$, f’(x) = 30.$\frac{2}{5}$ – 14 = 12 – 14 = – 2 < 0

At x = $\frac{5}{2}$, f’(x) = 30.$\frac{5}{2}$ – 14 = 75 – 14 = 61 > 0

So, given function is increasing at x = $\frac{5}{2}$ and decreasing at x = $\frac{2}{5}$

(ii) Show that the function f(x) = 2x³ – 24x + 15 is increasing at x=2 and decreasing at x==$\frac{3}{2}$.

Solution:

Given f(x) = 2x³ – 24x + 15

f'(x) = 6x² – 24.

At x = 3, f’(x) = 6.3² – 24 = 12 – 14 = 54 – 24 = 30 > 0.

At x = $\frac{3}{2}$, f’(x) = 6.${\left( {\frac{3}{2}} \right)^2}$ – 24 = $\frac{{27 - 48}}{2}$ = $ - \frac{{21}}{2}$< 0

So, given function is increasing at x = 3 and decreasing at x = $\frac{3}{2}$

 

2. (i) Test whether the function f(x) =2x² – 4x + 3 is increasing or decreasing on both interval (1,4].

Solution:

Given f(x) = 2x² – 4x + 3

f’(x) = 4x – 4

For every value in between,

(1,4], f’(x) > 0.

So, Given function is increasing on the interval (1,4].

(ii) Examine whether the function f(x) = 16x – $\frac{{4{{\rm{x}}^3}}}{3}$ is increasing or decreasing on the interval (–∞,–2).

Solution:

Given f(x) = 16x – $\frac{{4{{\rm{x}}^3}}}{3}$

f’(x) = 16 – 4x².

For every value (–∞,–2), the function f’(x) < 0.

So, Given function is decreasing on the interval (–∞,–2).

(iii) Show that the function f(x) = –x³ + 6x² – 13x + 20 is decreasing for all x ϵ R.

Solution:

Given f(x) = –x³ + 6x² – 13x + 20

f’(x) = – 3x² + 12x – 13

= – 3x² + 12x – 12 – 1 = –3(x² – 4x + 4) – 1

= –3(x – 2)² – 1 is always negative for every values of x ϵ R.

So, Given function is decreasing for all x ϵ R. 

(iv) Show that the function 4x – $\frac{9}{{\rm{x}}}$ + 6 is increasing for all x ϵ R except x = 0.

Solution:

Given f(x) = 4x – $\frac{9}{{\rm{x}}}$ + 6.

f’(x) = 4 + $\frac{9}{{{{\rm{x}}^2}}}$

f’(x) = 4 + $\frac{9}{{{{\rm{x}}^2}}}$  is always positive except x = 0.

So, therefore function is increasing for all x ϵ R except x = 0.

 

3. Find the interval in which the following function are increasing or decreasing.

(i) f(x) = 3x² – 6x + 5

Solution:

Given f(x) = 3x² – 6x + 5

f’(x) = 6x – 6

Let 6x – 6 = 0 ⇒ x = 1

When x > 1,   f’(x) > 0

When x < 1, f’(x) < 0

So, given function is increasing for (1,∞) and decreasing for (–∞.1).

(ii) f(x) = x⁴ – $\frac{1}{3}$x³

Solution:

Given f(x) = x⁴ – $\frac{1}{3}$x³

f’(x) = 4x³ – x².

Let 4x³ – x² = 0 ⇒ x²(4x – 1) = 0

When x  = 0, or x = $\frac{1}{4}$.

When x < 0, f’(x) < 0.

When 0 < x <$\frac{1}{4}$,   f’(x) < 0

When x >$\frac{1}{4}$, f’(x) > 0

So, given function is decreasing for $\left( { - {\rm{\infty }},\frac{1}{4}} \right)$ and increasing for $\left( {\frac{1}{4},{\rm{\infty }}} \right)$.

(iii) f(x) = 5x³-135x+22

Solution:

f(x) = 5x³-135x+22

f’(x )=15x² – 135

Let 15x² – 135 = 0 ⇒ x² = 9 ⇒ x = ±3

When x <–3, f’(x) > 0

When –3 < x < 3,   f’(x) < 0

When x > 3, f’(x) > 0

So, given function is increasing for $\left( { - {\rm{\infty }},3} \right){\rm{U}}\left( {3,{\rm{\infty }}} \right)$ and decreasing for $\left( { - 3,3} \right)$ 

(iv) f(x) = 6 + 12x + 3x² – 2x²

Solution:

f(x) = 6 + 12x + 3x² – 2x²

f'(x) = 12 + 6x – 6x²

Let  12 + 6x – 6x² = 0

Or, x² – x – 2 = 0

Or, x² – 2x + x – 2 = 0

Or, x(x – 2) + 1 (x – 2) = 0

Or, (x – 2)(x + 1) = 0

Either, x = 2 or , x = –1

When x <–1, f’(x) < 0

When –1 < x < 2,   f’(x) > 0

When x > 2, f’(x) < 0

So, given function is decreasing for $\left( { - {\rm{\infty }}, - 1} \right){\rm{U}}\left( {2,{\rm{\infty }}} \right)$ and increasing for $\left( { - 1,2} \right)$.

(v) f(x) = x³ – 12x

Solution:

f(x) = x³ – 12x

f'(x) = 3x² – 12

Let  3x² – 12 = 0 ⇒ x² = 4 ⇒ x= ± 2.

When x <–2, f’(x) > 0.

When –2 < x < 2,   f’(x) < 0

When x > 2, f’(x) > 0.

So, given function is increasing for $\left[ { - 3, - 2} \right){\rm{U}}\left( {2,5} \right]$ and decreasing for $\left( { - 2,2} \right)$.

 

4. Find the absolute maximum and the absolute minimum values of the following function on the given intervals:

(i) f(x) = 3x² – 6x + 4 on [-1,2]

Solution:

Given f(x) = 3x² – 6x + 4

f'(x) = 6x – 6

For, maximum or minimum f’(x) = 0

i.e. 6x – 6 = 0 ⇒ x = 1

When x = –1, f(x) = 3(–1)²– 6(–1) + 4 = 13

When, x = 2, f(x) = 3.2² – 6.2 + 4 = 4

When x = 1, f(x) = 3.1² – 6.1 + 4 = 1.

So, absolute maximum = 13, Absolute minimum = 1. 

(ii) f(x) = 2x³ – 9x² on [-2,4]

Solution:

Given f(x) = 2x³ – 9x²

f'(x) = 6x² – 18x

For, maximum or minimum f’(x) = 0

i.e. 6x – 18x = 0

or, 6x(x – 3) = 0

When x = 0, or, x = 3.

When, x = –2, f(x) = 2.(–2)³ – 9*(–2)² = – 52

When x = 4, f(x) = 2.4³ – 9.4² = – 16.

When x = 0, f(x) = 0

When x = 3, f(x) = 2.3³ – 9.3² = – 27.

So, absolute maximum = 0, Absolute minimum = –52. 

(iii) f(x) = x³-6x²+9x on [0,5]

Solution:

Given: f(x) = x³-6x²+9x

f’(x) = 3x² – 12x + 9

For, maximum or minimum f’(x) = 0

3x² – 12x + 9 = 0

or, x² – 4x + 3 = 0.

Either, x = 1 or x = 3,

When x = 0, f(x) = 0³ – 6.0² + 9.0 = 0

When, x = 3, f(x) = 3³ – 6.3² + 9.3 = 0

When x = 1, f(x) = 1³ – 6.1² + 9.1 = 4.

When x = 5, f(x) = 5³ – 6.5² + 9.5 = 20

When x = 3, f(x) = 2.3³ – 9.3² = – 27.

So, absolute maximum = 20, Absolute minimum = 0. 

(iv) f(x)=2x³-15x²+36x+10 on [1,4]

Solution:

Given: f(x)=2x³-15x²+36x+10

f’(x) = 6x² – 30x + 36

For, maximum or minimum f’(x) = 0

6x² – 30x + 36 = 0

or, x² – 5x + 6 = 0

or, (x – 1)(x – 5) = 0.

Either, x = 1 or x = 5,

When x = 1, f(x) = 2.1³ – 15.1³ + 36.1 + 10 = 33.

When, x = 4, f(x) = 2.4³ – 15.4² + 36.4 + 10 = 42.

Since, 5 ∉$\left[ {1,4} \right]$.

So, absolute maximum = 42, Absolute minimum = 33.

 

5. Find the local maxima and local minima and the point of inflection

(i) f(x) = 3x² – 6x + 3

Solution:

f(x) = 3x² – 6x + 3

f’(x) = 6x – 6.

f'’(x) = 6.

For maxima and minima, f’(x) = 0.

6x – 6 = 0

Or, 6x = 6 ⇒ x = 1.

At x = 1, f’’(x) = 6 > 0.

So, f(x) has minimum value at x =1.

Min. value = f(1) = 3 * (1)² – 6 * 1 + 3 = 0

For all x, f’’(x) = 6 ≠ 0, so there is no point of inflection. 

(ii) f(x) = x³ – 12x + 8

Solution:

f(x) = x³ – 12x + 8

f’(x) = 3x² – 12.

f'’(x) = 6x.

For maxima and minima, f’(x) = 0.

3x² – 12 = 0.

3x² = 12

Or, x² = 3 –> x = ± 2.

At x = 2, f’’(x) = 6 * 2 = 12 > 0.

So, f(x) has minimum value at x = 2.

Min. value of f(2) = f(2) = (2)³ – 12*2 + 8 = –8.

Again, x = –2, f’’(x) = 6 * (–2) = –12 < 0.

So, f(x) has maximum value at x = –2.

Max. value of f(x) = f(–2) = (–2)³ – 12 * (–2) + 8 = 24.

f'(x) = 0 if 6x = 0 ⇒ x = 0

So, x = 0 is the point of inflection.

(iii) f(x) = x³ – 6x² + 3

Solution:

f(x) = x³ – 6x² + 3

f’(x) = 3x² – 12x.

f'’(x) = 6x – 12.

For maxima and minima, f’(x) = 0.

3x² – 12x = 0.

3x(x – 4) = 0.

Either, 3x = 0 – ⇒ x = 0

Or, x – 4  ⇒ x = 4.

At x = 4, f’’(x) = 6 * 4 – 12 = 12 > 0.

So, f(x) has minimum value at x = 4.

Min. value = (4)³ – 6(4)² + 3 = –29.

f'’(x) = 0 if 6x – 12 = 0

6x = 12  ⇒ x = 2

So, x = 2 is the point of inflection. 

(iv) f(x) = 2x³ – 15x² + 36x + 5

Solution:

f(x) = 2x³ – 15x² + 36x + 5

f’(x) = 6x² – 30x + 36

f'’(x) = 12x – 30.

For maxima and minima, f’(x) = 0.

Or, 6x² – 30x + 36 = 0

Or, x² – 5x + 6 = 0

Or, (x – 2)(x – 3) = 0

Either, x – 2 = 0, x ⇒ 2

Or, x – 3 = 0, x ⇒ 3.

At x = 2, f’’(x) = 12*2 – 30 = –6 > 0.

So, f(x) has maximum value at x = 2.

Max. value = 2.(3)³ – 15(2)² + 36 * 2 + 5 = 33

At x = 3, f’’(x) = 12 * 3 – 30 = 6 > 0.

So, f(x) has minimum value at x = 3.

Min. value = 2(3)² – 15(3)² + 36 * 3 + 5 = 32

f’’(x) = 0  if 12x – 30 = 0

12x = 30  ⇒ x = $\frac{{30}}{{12}}$ = $\frac{5}{2}$.

So, x = $\frac{5}{2}$ is the point of inflection.

(v) f(x) = 2x³ – 9x² – 24x + 3

Solution:

f(x) = 2x³ – 9x² – 24x + 3

f’(x) = 6x² – 18x – 24.

f'’(x) = 12x – 18.

For maxima and minima, f’(x) = 0.

Or, 6x² – 18x – 24. = 0

Or, x² – 3x + 4 = 0

Or, (x – 4)(x + 1) = 0

Either, x – 4 = 0, x ⇒ 4

Or, x – 1 = 0, x ⇒ –1.

At x = 4, f’’(x) = 12*4 – 18 = 30 > 0.

So, f(x) has minimum value at x = 4.

Min. value = 2.(4)³ – 9(4)² – 24(4)+ 3 = –109

At x = –1, f’’(x) = 12 * (–1) – 18 = –30 < 0.

So, f(x) has maximum value at x = –1.

Max. value = 2(–1)³ – 9(–1)² – 24(–1) + 3 = 16.

f’’(x) = 0  if 12x – 18 = 0

12x = 18  ⇒ x = $\frac{{18}}{{12}}$ = $\frac{3}{2}$.

So, x = $\frac{3}{2}$ is the point of inflection

(vi) f(x) = 4x³ – 15x² + 12x + 7

Solution:

f(x) = 4x³ – 15x² + 12x + 7

f’(x) = 12x² – 30x + 12.

f'’(x) = 24x – 30.

For maxima and minima, f’(x) = 0.

Or, 12x² – 30x + 12 = 0

Or, 2x² – 5x + 2 = 0

Or, 2x² – 4x – x + 2 = 0

Or, (x – 2)(2x – 1) = 0

Either, x – 2 = 0, x ⇒ 2

Or, 2x – 1 = 0, x ⇒ $\frac{1}{2}$.

At x = 2, f’’(x) = 24*2 – 30 = 18 > 0.

So, f(x) has minimum value at x = 2.

Min. value = 4.(2)³ – 15(2)² + 12(2)+ 7 = 3

At x = $\frac{1}{2}$, f’’(x) = 24 * $\frac{1}{2}$– 30 = –18 < 0.

So, f(x) has maximum value at x = $\frac{1}{2}$.

Max. value = 4.${\left( {\frac{1}{2}} \right)^3} - 15.{\left( {\frac{1}{2}} \right)^2} + 12.\left( {\frac{1}{2}} \right) + 7$ = $\frac{{39}}{4}$.

f’’(x) = 0  if 24x – 30 = 0

24x = 30  ⇒ x = $\frac{5}{4}$.

So, x = $\frac{5}{4}$ is the point of inflection. 

(vii) f(x) = 4x³ – 6x² – 9x + 1 on the interval (-1,2)

Solution:

f(x) = 4x³ – 6x² – 9x + 1

f’(x) = 12x² – 12x – 9,

f'’(x) = 24x – 12.

For maxima and minima, f’(x) = 0.

Or, 12x² – 12x – 9 = 0

Or, 4x² – 4x – 3 = 0

Or, 4x² – 6x + 2x – 3 = 0

Or, 2x(2x – 3) + 1(2x – 3) = 0

Or, (2x – 3)(2x + 1) = 0

Either, x = $\frac{3}{2}$, or x = $ - \frac{1}{2}$.

When x = $\frac{3}{2}$, f’’(x) = 24*$\frac{3}{2}$ – 12= 24> 0.(case of minima)

So, given function has minimafor x = $\frac{3}{2}$ and its local minimum value is f$\left( {\frac{3}{2}} \right)$ = $4.{\left( {\frac{3}{2}} \right)^2} - 6.{\left( {\frac{3}{2}} \right)^2} - 9.\left( {\frac{3}{2}} \right) + 1$.

= 4 * $\frac{{27}}{8}$ – 6 * $\frac{9}{4}$ – $\frac{{27}}{2}$ + 1 = $ - \frac{{25}}{2}$.

when x = $ - \frac{1}{2}$, f’’(x) = 24 * $\left( { - \frac{1}{2}} \right)$– 12 = –24< 0.(case of maxima)

So, given function has maxima for x = $ - \frac{1}{2}$ and its local maximum value is f$\left( { - \frac{1}{2}} \right)$ = $4.{\left( { - \frac{1}{2}} \right)^3} - 6.{\left( { - \frac{1}{2}} \right)^2} - 9.\left( { - \frac{1}{2}} \right) + 1$.

= $ - \frac{1}{2} - \frac{3}{2} + \frac{9}{2}$ + 1 = $\frac{7}{2}$

For point of inflection, f’’(x) = 0.

So, 24x – 12 = 0 ⇒ x = $\frac{1}{2}$. 

(viii) f(x) = x + $\frac{{100}}{{\rm{x}}}$ – 5

Solution:

f(x) = x + $\frac{{100}}{{\rm{x}}}$ – 5

f’(x) = 1 – $\frac{{100}}{{{{\rm{x}}^2}}}$

f'’(x) = $\frac{{200}}{{{{\rm{x}}^3}}}$

For maxima and minima, f’(x) = 0.

Or, 1 – $\frac{{100}}{{{{\rm{x}}^2}}}$ = 0

Or, x² = 100

Or, x = ± 10.

At x = 10, f’’(x) = $\frac{{200}}{{{{10}^3}}}$= $\frac{1}{5}$> 0.

So, f(x) has minimum value at x = 10.

Min.value = 10 + $\frac{{100}}{{10}}$ – 5 = 15

At x = – 10, f’’(x) = $\frac{{200}}{{{{\left( { - 10} \right)}^3}}}$= $ - \frac{1}{5}$< 0

So, f(x) has maximum value at x = –10.

Or, Max. value = – 10 + $\frac{{100}}{{ - 10}}$ – 5 = –25.

No. finite value of x makes f’’(x) = 0,so there is no point of inflection.

 

6. Show that the following function have neither maximum nor minimum values.

(i) f(x) = x³ – 6x² + 24x + 4

Solution:

f(x) = x³ – 6x² + 24x + 4

f’(x) = 3x² – 12x + 24

= 3{x² – 4x + 8} = 3{x² – 4x + 4 + 4}

= 3{(x – 2)² + (2)²} ≠ 0 for all real values of x.

So, f(x) has neither maximum nor minimum value. 

(ii) f(x) = x³ – 6x² + 12x – 3
Solution:

f(x) = x³ – 6x² + 12x – 3

f’(x) = 3x² – 12x + 12.

f’’(x) = 6x – 12.

For maxima or minima, f’(x) = 0.

Or, 3x² – 12x + 12 = 0

Or, x² – 4x + 4 = 0

Or, (x – 2)² = 0

Or, (x – 2) = 0

So, x = 2.

At, x = 2,  f’(x) =  6 * 2 – 12 = 0

At x = 2, f’’(x) = 6 ≠ 0.

So, f(x) has neither maximum nor minimum value.

 

7. Determine where the graph is concave upward and where it is concave downwards of the following function:

(i) f(x) = x⁴ – 2x³ + 5

f(x) = x⁴ – 2x³ + 5

f’(x) = 4x³ – 6x²

f’’(x) = 12x² – 12x = 12x(x – 1)

For, x > 1,   f’’(x) > 0, so the graph is concave upwards.

For, x < 0. f'’(x) > 0, so the graph is concave upwards.

For  0 < x < 1, f’’(x) < 0, so the graph is concave downwards. 

(ii) f(x) = x⁴ – 8x³ + 18x² – 24

Solution:

f(x) = x⁴ – 8x³ + 18x² – 24

f’(x) = 4x³ – 24x² + 36x

f’’(x) = 12x² – 48x + 36. = 12(x² – 4x + 3) = 12(x – 3)(x – 1).

For, x > 3,   f’’(x) > 0, so the graph is concave upwards.

For, x < 1. f'’(x) > 0, so the graph is concave upwards.

For  1 < x < 3, f’’(x) < 0, so the graph is concave downwards. 

(iii) y = 3x⁵ + 10x³ + 15x

Solution:

y = 3x⁵ + 10x³ + 15x

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 15x⁴ + 30x³ + 15.

Or, $\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = 60x³ + 60x = 60x(x² + 1).

For, x > 0,   f’’(x) < 0, so the graph is concave upwards.

For, x < 0. f'’(x) > 0, so the graph is concave downwards.

(iv) f(x) = x³ – 9x² defined on [-2,5]

Solution:

f(x) = x³ – 9x²

f’(x) = 3x² – 18x

f’’(x) = 6x – 18.

Let f’’(x) = 6x – 18 = 0 ⇒ x = 3.

When –2 ≤ x < 3,  f’’(x) < 0.

When 3 < x ≤ 5,  f’’(x) > 0

So, given function is concave upward fir 3 < x ≤ 5 and downward for –2 ≤ x < 3.

 

8. A man who has 144 meters of fencing materials wishes to enclose a rectangular garden. Find the maximum area he can encloses.

Solution:
Let x = length, y = breadth and A = area of the rectangular garden.
Perimeter = 144

Or, 2x + 2y = 144

Or, x + y = 72.

So, y = 7x – x …(i)

A = x.y = x(72 – x) = 72x – x².

Or, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 72 – 2x, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = –2.

For maxima or minima, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 0

Or, 72 – 2x = 0

Or, 2x = 72

So, x = 36

When x = 36, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = –2 < 0, so A has maximum value,

When x = 36, y = 72 – x = 72 – 36 = 36.

Max.area = x.y = 36 * 36 = 1296m².

 

9. Show that the rectangle of largest possible are, for a given perimeter is square.

Solution:

Let x = length, y = breadth and A = area of the rectangular garden.
Perimeter = 2a(suppose)

Or, 2x + 2y = 2a

Or, x + y = a.

So, y = a – x….(i)

A = x.y = x(a – x) = ax – x².

Or, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = a – 2x, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = –2.

For maxima or minima, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 0

Or, a – 2x = 0

So, x =$\frac{{\rm{a}}}{2}$.

When x = $\frac{{\rm{a}}}{2}$, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = –2 < 0, A has max.value.

When x = a/2,  y = a – x = a – a/2 = a/2.

So, x = y i.e. length = breadth, hence the rectangle is a square for max. area.

 

10. A window is in the form of rectangle surmounted by a semi-circle. If the total perimeter is 9 meters, find the radius of semi-circle for the greatest window area.

Solution:

Let 2x = length, y = breadth of the window in the form a rectangle.

Then the radius of the semi – circle on the rectangle = x.

Perimeter = 2x + 2y + $\frac{1}{2}$2πx

Or, 9 = 2x + 2y

So, y = $\frac{1}{2}\left( {9 - 2{\rm{x}} - {\rm{\pi x}}} \right)$.

A = area of the window = 2x.y + $\frac{1}{2}$π.x².

= 2x.$\frac{1}{2}$(9 – 2x – πx) + $\frac{1}{2}$πx².

= 9x – 2x² – πx² + $\frac{1}{2}$πx².

= 9x – 2x² – $\frac{1}{2}$πx²

Or, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 9 – 4x – πx, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = – 4 – π

For, maxima or minima, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 0.

Or, 9 – 4x – πx = 0

Or, (4 + π)x = 9

So, x = $\frac{9}{{4 + {\rm{\pi }}}}$.

When x = $\frac{9}{{4 + {\rm{\pi }}}}$, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = – 4 – π < 0 so A has maximum value.
So, x = radius of the semi – circle = $\frac{9}{{4 + {\rm{\pi }}}}$ meter.

 

11. A closed cylinder can is to be made so that its volume is 52cm³. Find its dimension if the surface is to be minimum.

Solution:

Let r = Radius, h = Height and V = volume of the cylindrical can.

V = πr²h.

Or, 52 = πr²h

So, h = $\frac{{52}}{{{\rm{\pi }}{{\rm{r}}^2}}}$ ….(i)

Again, S = 2πrh + 2πr², where S = surface area.

= 2πr. $\frac{{52}}{{{\rm{\pi }}{{\rm{r}}^2}}}$ + 2πr² = $\frac{{104}}{{\rm{r}}}$ + 2πr²

Or, $\frac{{{\rm{dS}}}}{{{\rm{dr}}}}$ = $ - \frac{{104}}{{{4^2}}} + 4{\rm{\pi r}}$

Or, $\frac{{{{\rm{d}}^2}{\rm{S}}}}{{{\rm{d}}{{\rm{r}}^2}}}$ = $\frac{{208}}{{{{\rm{r}}^3}}}$ + 4π.

For maxima or minima, $\frac{{{\rm{dS}}}}{{{\rm{dr}}}}$ = 0.

Or, $ - \frac{{104}}{{{{\rm{r}}^2}}}$ + 4πr = 0

Or, 4πr³ = 104

Or, r³ = $\frac{{104}}{{4{\rm{\pi }}}}$ = $\frac{{26}}{{\rm{\pi }}}$.

So, r = $\sqrt[3]{{\frac{{26}}{{\rm{\pi }}}}}$

When r = $\sqrt[3]{{\frac{{26}}{{\rm{\pi }}}}}$, $\frac{{{{\rm{d}}^2}{\rm{S}}}}{{{\rm{d}}{{\rm{r}}^2}}}$ = $\frac{{208}}{{{{\rm{r}}^3}}}$ + 4π = $\frac{{208}}{{\frac{{26}}{{\rm{\pi }}}}}$ + 4π = 12π.

So, S has minimum value when r = $\sqrt[3]{{\frac{{26}}{{\rm{\pi }}}}}$

From(i), h = $\frac{{52}}{{{\rm{\pi }}{{\rm{r}}^2}}} = \frac{{52{\rm{r}}}}{{{\rm{\pi }}{{\rm{r}}^3}}} = \frac{{52{\rm{r}}}}{{26}}$ = 2r.

i.e. h : r = 2 : 1, where r = $\sqrt[3]{{\frac{{26}}{{\rm{\pi }}}}}$.

 

12. A gardener having 120m of fencing wishes to enclose a rectangular plot of the land and also to erect a fence across the land parallel to two of the sides. Find the maximum area he can enclose.

Solution:

Let length = x, breadth = y and A = Area of the rectangle plot of land.

2x + 2y + x = 120

3x + 2y = 120

Y = $\frac{1}{2}$(120 – 3x)…..(i)

A = x.y = x. $\frac{1}{2}$ (120 – 3x)

= $\frac{1}{2}$(120x – 3x²)

Or, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}$(120 – 6x) = 60 – 3x

Or, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = –3

For maximum or minima, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}}$ = 0

Or, 60 – 3x = 0

So, x = 20.

When x = 20, $\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = – 3 < 0, so A has maximum value.

From(i), y = $\frac{1}{2}$.(120 – 3*20) = 30.

Max.area = x.y = 20.30 = 600m².

 

13. Find two number whose sum is 10 and the sum of whose square is mimimum.

Solution:

Let x and y be two numbers. Then,

X + y = 10

Y = 10 – x ….(i)

Let S = x² + y² = x² + (10 – x)²

S = x² + 100 – 20x + x² = 2x² – 20x + 100

Or, $\frac{{{\rm{dS}}}}{{{\rm{dx}}}}$ = 4x – 20, $\frac{{{{\rm{d}}^2}{\rm{S}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = 4.

For maxima or minima, $\frac{{{\rm{dS}}}}{{{\rm{dx}}}}$ = 0

Or, 4x – 20 = 0

So, x = 5,

When x = 5, $\frac{{{{\rm{d}}^2}{\rm{S}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = 4 > 0, so S has maximum value.

When x = 5, y = 10 – x = 10 – 5 = 5.

So, the two numbers are 5 and 5.