{"id":22164,"date":"2024-01-11T12:54:04","date_gmt":"2024-01-11T07:24:04","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22164"},"modified":"2024-01-12T16:16:15","modified_gmt":"2024-01-12T10:46:15","slug":"inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10h","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10h\/","title":{"rendered":"Inter 1st Year Maths 1B Applications of Derivatives Solutions Ex 10(h)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the points of local extrema (if any) and local extrema of the following functions each of whose domain is shown against the function.
\ni) f(x) – x\u00b2, \u2200 x \u2208 R.
\nSolution:
\nf(x) = x\u00b2
\nf'(x) = 2x \u21d2 f”(x) = 2
\nFor maximum or minimum f'(x) = 0
\n2x = 0
\nx = 0
\nNow f”(x) = 2 > 0 ,
\n\u2234 f(x) has minimum at x = 0
\nPoint of local minimum x = 0
\nLocal minimum = 0.<\/p>\n

ii) f(x) = sin x, [0, 4\u03c0)
\nSolution:
\nGiven f(x) = sinx
\n\u21d2 f'(x) = cosx
\n\u21d2 f”(x) = -sinx
\nFor max on min,
\nf'(x) = 0
\ncos x = 0
\n\u21d2 x = \\(\\frac{\\pi}{2},\\frac{3\\pi}{2},\\frac{5\\pi}{2},\\frac{7\\pi}{2}\\)<\/p>\n

i) f”(\\(\\frac{\\pi}{2}\\)) = -sin\\(\\frac{\\pi}{2}\\) = -1 < 0
\nf(x) = sin \\(\\frac{\\pi}{2}\\) = 1
\n\u2234 Point of local maximum x = \\(\\frac{\\pi}{2}\\)
\nlocal maximum – 1<\/p>\n

ii) f”(\\(\\frac{3\\pi}{2}\\)) = – sin \\(\\frac{3\\pi}{2}\\) = -1 > 0
\nf(x) = sin \\(\\frac{3\\pi}{2}\\) = -1
\n\u2234 Point of local minimum x = \\(\\frac{3\\pi}{2}\\)
\nlocal minimum x = -1<\/p>\n

iii) f”(\\(\\frac{5\\pi}{2}\\)) = – sin \\(\\frac{5\\pi}{2}\\) = -1 > 0
\nf(x) = sin \\(\\frac{5\\pi}{2}\\) = 1
\n\u2234 Point of local maximum x = \\(\\frac{5\\pi}{2}\\)
\nlocal maximum = 1<\/p>\n

iv) f”(\\(\\frac{7\\pi}{2}\\)) = – sin \\(\\frac{7\\pi}{2}\\) = -1 > 0
\nf(x) = sin \\(\\frac{7\\pi}{2}\\) = -1
\n\u2234 Point of local maximum x = \\(\\frac{7\\pi}{2}\\)
\nlocal maximum = 1<\/p>\n

\"Inter<\/p>\n

iii) f (x) = x\u00b3 – 6x\u00b2 + 9x + 15 \u2200 x \u2208 R.
\nSolution:
\nf (x) = 3x\u00b2 – 12x + 9 and f'(x) = 6x – 12
\nFor maximum or minimum f(x) = 0
\n\u21d2 3x\u00b2 – 12x + 9 = 0
\n\u21d2 x\u00b2 – 4x + 3 = 0
\n\u21d2 (x – 1) (x – 3) = 0
\n\u21d2 x = 1 or 3
\nNow f”(1) = 6(1) – 12 = – 6 < 0
\n\u2234 f(x) has maximum value at x = 1
\nMax. valueis f(1)= 1\u00b3 – 6(1)\u00b2 + 9(1) + 15
\n= 1 – 6 + 9 + 15 = 19<\/p>\n

f”(3) = 6(3) – 12 = 18 – 12 = 6 > 0
\n\u2234 f(x) has minimum value at x = 3
\nMin. value is f(3) = 33 – 6.32 + 9.3 + 15
\n= 27 – 54 + 27+ 15
\n= 15<\/p>\n

iv) f(x) = \\(\\sqrt{1-x}\\) \u2200 x \u2208 (0, 1)
\nSolution:
\n\"Inter
\nFor max. or min. f'(x) = 0
\n\"Inter<\/p>\n

v) f(x) = 1\/x\u00b2 + 2 \u2200 x \u2208 R
\nSolution:
\n\"Inter
\n\u2234 For maximum or minimum f (x) = 0
\n\u21d2 \\(\\frac{-2x}{(x^{2}+2)^{2}}\\) = 0 \u21d2 x = 0
\nf”(0) = \\(\\frac{2(0-2)}{(0+2^{3})}=\\frac{-4}{8}=\\frac{-1}{2}\\) < 0
\n\u2234 f(x) has max. value at x = 0
\nMax. value is f(0) = \\(\\frac{1}{0+2}=\\frac{1}{2}\\)<\/p>\n

vi) f(x) = x\u00b2- 3x \u2200 x \u2208 R
\nSolution:
\nf(x) = 3x\u00b2 – 3 and f”(x) = 6x
\n\u2234 For maximum or minimum f(x) = 0
\n\u21d2 3x\u00b2 – 3 = 0
\n\u21d2 x\u00b2 – 1 = 0
\n\u21d2 x = \u00b1 1
\nNow f”(1) = 6(1) = 6 > 0
\n\u2234 f(x) has minimum at x = 1
\nMinimum value is f(1) = 1\u00b3 – 3(1) = -2
\nf”(-1) = 6(-1) = -6 < 0
\n\u2234 f(x) has maximum value at x = -1
\nMaximum value is f(-1) = (-1)\u00b3 – 3(-1)
\n= -1 + 3 = 2<\/p>\n

vii) f(x) = (x -1) (x + 2)\u00b2 \u2200 x \u2208 R
\nSolution:
\nf(x) = (x – 1) (x + 2)\u00b2
\nf(x) = (x – 1) 2(x + 2) + (x + 2)\u00b2
\n= 2(x – 1) (x + 2) + (x + 2)\u00b2
\nf”(x) = 2(x – 1) + 2(x + 2) + 2(x + 2)
\n= 2(3x + 3) = 6(x + 1)
\n\u2234 For maximum or minimum f'(x) = 0
\n2(x – 1) (x + 2) + (x + 2)\u00b2 = 0
\n(x + 2) [2(x – 1) + (x + 2)] = 0
\n\u21d2 (x + 2) (3x) = 0
\n\u21d2 x = 0, x = -2
\nNow f”(0) = 6(0 + 1) = 6 > 0
\n\u2234 f(x) has min. value at x = 0
\nMin. value is f(0) = (0 – 1) (0 + 2)\u00b2 = -4
\nf'(-2) = 6 (-2 + 1) = -6 < 0
\n\u2234 f(x) has max. value at x = -2
\nMax. value is f(-2) = (-2 -1) (-2 + 2)\u00b2 = 0<\/p>\n

viii) f(x) = \\(\\frac{x}{2}+\\frac{2}{x}\\) \u2200 x \u2208 (0, \u221e)
\nSolution:
\nf'(x) = \\(\\frac{1}{2}-\\frac{2}{x^{2}}\\) and f”(x) = \\(\\frac{4}{x^{3}}\\)
\n\u2234 For max. or min. f'(x) = 0
\n\u21d2 \\(\\frac{1}{2}-\\frac{2}{x^{2}}\\) = 0 \u21d2 x\u00b2 – 4 = 0 \u21d2 x = \u00b1 2
\nf”(2) = \\(\\frac{4}{2^{3}}\\) = \\(\\frac{1}{2}\\) > 0 (Since x > 0)
\n\u2234 f(x) has min. value at x = 2
\nMin. value is f(2) = \\(\\frac{2}{2}+\\frac{2}{2}\\) = 1 + 1 = 2.<\/p>\n

ix) f(x) = – (x – 1)\u00b3 (x + 1)\u00b2 \u2200 x \u2208 R
\nSolution:
\nf(x) = -(x – 1)\u00b3 (x + 1)\u00b2 = (1 – x)\u00b3 (x + 1)\u00b2
\nf”(x) = (1 – x)\u00b3 2(x + 1) + 3(1 – x)\u00b2 (-1) (x + 1)\u00b2
\n= (1 – x)\u00b2 (x + 1) {2(1 – x) – 3(x + 1)}
\n= (1 – x)\u00b2 (x + 1) {2 – 2x – 3x – 3}
\n= (1 – x)\u00b2 (x + 1) (-1 – 5x)
\nf”(x) = (1 – x)\u00b2 (x + 1) (-5) + (1 – x)\u00b2 (-1 – 5x) + (x + 1) (-1 -5x) 2(1 – x) (-1)
\n= -5 (1 -x)\u00b2 (x+ 1) – (1 + 5x) (1 – x)\u00b2 + (x + 1) (1 + 5x) 2(1 – x)
\n\u2234 For maximum or minimum f(x) = 0
\n(1 – x)\u00b2 (x + 1) (-1 – 5x) = 0
\n\u21d2 x = \u00b1 1 or -1\/5
\nf”(1) = 0 – 0 + 0 \u21d2 critical value at x = 0
\nf”(1 +1)\u00b2 (-1) = 0 – (1 – 5) + 0 = 16 > 0
\n\u2234 f(x) has min. value at x = -1
\nMin. value = f(-1) = (1 + 1)\u00b3 (-1 + 1)\u00b2 = 0
\nf”(- \\(\\frac{1}{5}\\)) < o
\n\u21d2 f(x) has max. value at x = – \\(\\frac{1}{5}\\)
\nMin. value is f (-\\(\\frac{1}{5}\\)) = \\(\\frac{3456}{3125}\\)<\/p>\n

x) f(x) = x\u00b2 e3x<\/sup> \u2200 x \u2208 R
\nSolution:
\nf'(x) = x\u00b2 e3x<\/sup> .3 + e3x<\/sup>. 2x
\nFor maximum or minimum f'(x) = 0
\n3x\u00b2 e3x<\/sup> + 2e3x<\/sup>. x = 0
\nx\u00b2 e3x<\/sup> (3x + 2) = 0
\nx = 0, x = \\(\\frac{-2}{3}\\) and e = 0 is not possible
\nNow f”(x) = 3(x\u00b2 e3x<\/sup>. 3 + e3x<\/sup> 2x)
\n+ e3x<\/sup> 2 + 2x e3x<\/sup>
\nf”(x) = 9x\u00b2e3x<\/sup> + 6x
\ne3x<\/sup>+ 2 e3x<\/sup> + 6xe3x<\/sup>
\n= 9x\u00b2e3x<\/sup> + 12xe3x<\/sup> + 2e3x<\/sup>
\nf\u201d(0) = 2 > 0
\n\u2234 Point of local minimum = 0
\nlocal minimum = 0
\nf”(\\(\\frac{-2}{3}\\)) = \\(\\frac{-2}{e^{2}}\\) < 0
\n\u2234 point of local maximum = \\(\\frac{-2}{3}\\)
\nlocal maximum = \\(\\frac{4}{9e^{2}}\\)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nProve that the following functions do not have absoute maximum and absolute minimum.
\ni) ex<\/sup> in R
\nSolution:
\nf(x) = ex<\/sup> and f”(x) = ex<\/sup>
\n\u2234 For maxima or minima f'(x) = 0 \u21d2 ex<\/sup> = 0
\n\u21d2 x value is not defined
\nHence it has no maxima or minima.<\/p>\n

ii) log x in (0, \u221e)
\nSolution:
\nf(x) = \\(\\frac{1}{x}\\) and f”(x) = – \\(\\frac{1}{x^{2}}\\)
\nf (x) = 0 \u21d2 x value is not defined
\n\u21d2 f(x) has no maxima or minima.<\/p>\n

iii) x\u00b3 + x\u00b2 + x + 1 in R
\nSolution:
\nf (x) = 3x\u00b2 + 2x + 1 gives imaginary values.
\n\u21d2 It has no maximum or minimum values.<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the absolute maximum value and absoulte minimum value of the following functions on the domain specified against the function.
\nSolution:
\nf(x) = x\u00b3 on (-2, 2)
\nf(x) = 3x\u00b2 and f'(x) – 6x
\nthe value f(-2) = (-2)\u00b3 = – 8
\nMax Value f(2) = 2\u00b3 = 8<\/p>\n

ii) f(x) = (x – 1)\u00b2+ 3 on [-3, 1]
\nSolution:
\nf(x) = 2(x – 1) and f'(x) = 2
\nMax. value f(-3) = (-3 – 1)\u00b2 + 3 = 16 + 3 = 19
\nMin. value f(l) = 0 + 3 = 3<\/p>\n

iii) f(x) = 2|x| on [-1, 6]
\nSolution:
\nf'(x) = \\(\\frac{2|x|}{x}\\)
\nFor max. or min., f'(x) = 0
\n\\(\\frac{2|x|}{x}\\) = 0 \u21d2 x = 0
\nf(0) = 0
\nf(-1) – 2|-1| = 2
\nf(6) = 2(6) = 12
\nAbsolute minimum = 0
\nAbsolute maximum = 12<\/p>\n

iv) f(x) = sin x + cos x on [0, \u03c0]
\nSolution:
\nf'(x) cos x – sin x which exists at all x \u2208 (0, \u03c0)
\nNow f'(x) = 0 \u21d2 cos x – sin x = 0
\n\u21d2 tan x = 1
\n\u21d2 x = \\(\\frac{\\pi}{4}\\) \u2208 (0, \u03c0)
\nNow f(0) = sin 0 + cos 0 = 1
\nf(\\(\\frac{\\pi}{4}\\)) = sin \\(\\frac{\\pi}{4}\\) + cos \\(\\frac{\\pi}{4}\\)
\n= \\(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}\\)
\nf(\u03c0) = sin \u03c0 + cos \u03c0 = 0 – 1 = -1
\n\u2234 Minimum value : -1
\nMaximum value: \u221a2<\/p>\n

v) f(x) = x + sin 2x on [0, \u03c0]
\nSolution:
\nf(x) = x + sin 2x
\nf(x) = 1 + 2 cos 2x
\nf (x) = 0 \u21d2 2 cos 2x + 1 = 0
\n\u21d2 cos 2x = –\\(\\frac{1}{2}\\) = cos \\(\\frac{2 \\pi}{3}\\)
\n\u21d2 2x = \\(\\frac{2 \\pi}{3}\\)
\n\u21d2 x = \\(\\frac{\\pi}{3}\\) \u2208 (0, 2\u03c0)
\nNow f(0) = 0 + sin 2(0) = 0
\nf(\\(\\frac{\\pi}{3}\\)) = \\(\\frac{\\pi}{3}\\) + sin 2.\\(\\frac{\\pi}{3}\\) = \\(\\frac{\\pi}{3}+\\frac{\\sqrt{3}}{2}\\)
\nf(2\u03c0) = 2\u03c0 + sin 2. 2\u03c0 = 2\u03c0 + 0 = 2\u03c0
\nMinimum value = 0
\nMaximum value is = 2\u03c0<\/p>\n

Question 2.
\nUse the first derivative test to find local extrema of f(x) = x\u00b3 – 12x on R.
\nSolution:
\nf(x) = x\u00b3 – 12x
\nf'(x) = 3x\u00b2 – 12
\nf”(x) = 6x
\nFor maximum or minimum, f'(x) = 0
\n3x\u00b2 – 12 = 0
\n3x\u00b2 = 12
\nx = \u00b1 2
\nf”(2) = 12 > 0
\nPoint of local minimum at x = 2
\nLocal minimum = – 16
\nf”(-2) = -12 < 0
\nPoint of local maximum at x = -1
\nLocal maximum =16<\/p>\n

Question 3.
\nUse the first derivative test to find local extrema of f(x) = x\u00b2 – 6x + 8 on R.
\nSolution:
\nf(x) = x\u00b2 – 6x + 8
\nf\u2019(x) = 2x -6 \u21d2 f”(x) = 2
\nFor maximum or minimum f'(x) = 0
\n2x – 6 = 0
\n\u21d2 x = 3
\nf”(3) = 2 > 0
\n\u2234 Point of local minimum x = 3
\nLocal minimum = -1<\/p>\n

\"Inter<\/p>\n

Question 4.
\nUse the second derivative test to find local extrema of the function
\nf(x) =x\u00b3 – 9x\u00b2 – 48x + 72 on R.
\nSolution:
\nf(x) =x\u00b3 – 9x\u00b2 – 48x + 72.
\n\u21d2 f'(x) = 3x\u00b2 – 18x – 48
\n= 3(x – 8) (x + 2)
\nThus the stationary point are – 2 & 8
\nf\u201d(x) = 6x – 18 = 6(x – 3)
\nAt x= 8, f”(8) = 30 > 0
\n\u2234 f (8) = (8)\u00b3 – 9(8)\u00b2 – 48(8) + 72
\n= 512 – 576 – 384 + 72
\n= – 376
\nAt x= -2, f”(-2) = – 30 < 0
\nf(-2) = (-2)\u00b3 – 9(-2)\u00b2 – 48(-2)+72
\n= -8 – 36 + 96 + 72
\n= 124
\nLocal minimum = -376
\nLocal maximum = 124<\/p>\n

Question 5.
\nUse the second derivative test to find local extrema of the function. f(x) = -x\u00b3 + 12x\u00b2 – 5 on R.
\nSolution:
\nf(x) = -x\u00b3 + 12x\u00b2 – 5
\n\u21d2 f'(x) = -3x\u00b2 + 24x
\n= -3x(x – 8)
\nThus the stationary point are 0, 8
\nf”(x) = – 6x + 24
\nAt x= 0, f”(0) = 24 > 0.
\nf(0) = -5
\nAt x= 8, f”(8) = -24 < 0
\nf(8) = -8\u00b3 + 12(8)\u00b2 – 5
\n= -512 + 768 – 5 = 251
\nLocal minimum = -5
\nLocal maximum = 251<\/p>\n

Question 6.
\nFind local maximum or local minimum of f(x) = -sin 2x – x defined on [-\u03c0\/2, \u03c0\/2].
\nSolution:
\nf(x) =-sin 2x – x
\nf'(x) = -2cos 2x – 1
\nf”(x) = 4 sin 2x
\nThus the starting point are x = \\(\\frac{\\pi}{3},\\frac{-\\pi}{3}\\)
\n\"Inter
\nLocal minimum = –\\(\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{3}\\)
\nLocal maximum = \\(\\frac{\\sqrt{3}}{2}+\\frac{\\pi}{3}\\)<\/p>\n

Question 7.
\nFind the absolute maximum and absolute minimum of f (x) = 2x\u00b3 – 3x\u00b2 – 36x + 2 on the interval [0, 5].
\nSolution:
\nf(x) = 2x\u00b3 – 3x\u00b2 – 36x + 2
\nf(x) = 6x\u00b2 – 6x – 36
\nf(x) = 12x – 6
\nfor maxima or minima, f'(x) = 0
\n6x\u00b2 – 6x – 36 = 0
\nx\u00b2 – x – 6 = 0
\nx\u00b2 -3x + 2x – 6 = 0
\nx(x – 3) + 2(x – 3) = 0
\n(x + 2) (x – 3) = 0
\nx = 3, -2
\nf”(3) = 30 > 0
\nf(x) has max\/min value at x = 3
\nf(3) = 2(3)\u00b3 – 3(3)\u00b2 – 36(3) + 2
\n= 54 – 27 – 108 + 2
\n= -79
\nAbsolute minimum = – 79
\nSince 0 \u2264 x \u2264 5
\n\u2234 f(0) = 0 – 0 – 0 + 2
\n= 2
\n\u2234 Absolute maximum = 2.<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind the absolute extremum of f(x) = 4x – \\(\\frac{x^{2}}{2}\\) on [-2, \\(\\frac{9}{2}\\)]
\nSolution:
\nf(x) = 4x – \\(\\frac{x^{2}}{2}\\)
\nf'(x) =4 – x
\nf”(x) = – 1
\nfor maxima or minima f'(x) = 0
\n4 – x = 0
\nx = 4
\nf”(4) = -1 < 0
\n\u2234 f has maximum value at x = 4
\nf(4) = 16 – \\(\\frac{16}{2}\\) = 8
\nSince -2 \u2264 x \u2264 \\(\\frac{9}{2}\\)
\n\u2234 f(-2) = -8 – \\(\\frac{4}{2}\\)
\n= -8- 2 = -10
\n\u2234 Absolute minimum = -10
\nAbsolute maximum = 8<\/p>\n

Question 9.
\nFind the maximum profit that a company can make, if the profit function is given by P(x) = -41 + 72x – 18x\u00b2
\nSolution:
\nP(x) = – 41 + 72 x – 18x\u00b2.
\n\\(\\frac{dp(x)}{dx}\\) = 72 – 36x
\nfor maxima or minima, \\(\\frac{dp}{dx}\\) = 0
\n72 – 36x = 0
\nx = 2
\n\\(\\frac{d^{2}p}{dx^{2}}\\) = -36 < 0
\n\u2234 The profit f(x) is maximum for x = 2
\nThe maximum profit will be P(2) =
\n= -41 + 72(2) – 18(4)
\n= 31<\/p>\n

Question 10.
\nThe profit function P(x) of a company selling x items is given by P(x) = -x\u00b3 + 9x\u00b2 – 15x – 13 where x represents thousands of units. Find the absolute maximum profits if the company can manufacture a maximum of 6000 units.
\nSolution:
\nP(x) = -x\u00b3 + 9x\u00b2 – 15x – 13
\n\\(\\frac{dp(x)}{dx}\\) = -3x\u00b2 + 18x – 15
\nfor maximum or minimum \\(\\frac{dp}{dx}\\) = 0
\n-3x\u00b2 + 18x – 15 = 0
\nx\u00b2 – 6x + 5 = 0
\nx\u00b2 – 5x – x + 5 = 0
\nx(x- 5) – 1(x- 5) = 0
\n(x – 1) (x – 5) = 0
\nx = 1, 5
\nP(1) = -1 + 9 – 15 – 13 = -10
\nP(5) = -125 + 225 – 75 – 13 = 12
\n\u2234 Maximum profit = 12.<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nThe profit function P(x) of a company selling x items per day is given by P(x) = (150 – x) x – 1000. Find the number of items that the company should manufacture to get maximum profit. Also find the maximum profit.
\nSolution:
\nGiven that tbe profit function
\nP(x) = (150 – x)x -1000
\nfor maximum or minimum \\(\\frac{dp}{dx}\\) = 0
\n(150 – x(1) -x (-1) = 0
\n150 – 2x = 0
\nx = 75
\nNow \\(\\frac{d^{2}p}{dx^{2}}\\) = -2 < 0
\n\u2234 The profit P(x) is maximum for x = 75
\nThe company should sell 75 tems a day the maxma profit will be P (75) = 4625.<\/p>\n

Question 2.
\nFind the absolute maximum and absolute minimum of f(x) = 8x\u00b3 + 81x\u00b2 – 42x – 8 on [-8, 2].
\nSolution:
\nf(x) = 8x\u00b3 + 81x\u00b2 – 42x – 8
\nf'(x) = 24x\u00b2+ 162x – 42
\nFor maximum or minimum, f'(x) = 0
\n24x\u00b2 + 162x – 42 = 0
\n4x\u00b2 + 27x – 7 = 0
\n4x\u00b2 + 28x – x – 7 = 0
\n4x(x + 7) – 1(x + 7) = 0
\n(x + 7) (4x – 1) = 0
\nx = – 7 or \\(\\frac{1}{4}\\)<\/p>\n

f(- 8) = 8(-8)\u00b3 + 81(-8)\u00b2 – 42(-8) -8
\n= – 8(512) + 81(64) + 336 – 8
\n= -4096 + 5184 + 336 – 8
\n= 5520 – 4104
\n= 1416<\/p>\n

f(2) = 8(2)\u00b3 + 81 (2)\u00b2 – 42(2) – 8
\n= 64 + 324 – 84 – 8
\n= 296
\n\"Inter
\nf(-7) = 1246
\nAbsolute maximum = 1416
\nAbsolute minimum = \\(\\frac{-216}{16}\\)<\/p>\n

Question 3.
\nFind two positive integers whose sum is 16 and the sum of whose squares is minimum.
\nSolution:
\nSuppose x and y are the sum value
\nx + y =16
\n\u21d2 y = 16 – x
\nf(x) = x\u00b2 + y\u00b2 = x\u00b2 + (16 – x)\u00b2
\n= x\u00b2 + 256 + x\u00b2 – 32x
\nf'(x) = 4x – 32
\nfor maximum or minimum f'(x) = 0
\n\u21d2 4x – 32 = 0
\n4x = 32 x = 8
\nf”(x) = 4 > 0
\n\u2234 f(x) is minimum when x = 8
\ny = 16 – x = 16 – 8 = 9
\n\u2234 Required number are 8, 8.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind two positive integers x and y such that x + y = 60 and xy3 is maximum.
\nSolution:
\nx + y = 60 \u21d2 y = 60 – x ………… (1)
\nLet p = xy\u00b3 = x(60 – x)\u00b3
\n\\(\\frac{dp}{dx}\\) = x\u00b3(60 – x)\u00b2(-1) + (60-x)\u00b3
\n= -3x (60 – x)\u00b2 + (60 – x)\u00b3
\n= (60 – x)\u00b2 [-3x + 60 – x]
\n= (60 – x)\u00b2 (60 – 4x) = 4(60 – x)\u00b2 (15 – x)<\/p>\n

\\(\\frac{d^{2}p}{dx^{2}}\\) = 4[(60-x)\u00b2 (-1) + (15-x) 2(60-x) (-1)]
\n= 4(60 – x) [-60 + x – 30 + 2x]
\n= 4(60 – x) (3x – 90)
\n= 12(60 – x) (x – 30)
\nFor maximum or minimum \\(\\frac{dp}{dx}\\) = 0
\n\u21d2 4(60 – x)2 (15 -x) = 0
\n\u21d2 x = 60 or x = 15 ; x cannot be 60
\n\u2234 x = 15 \u21d2 y = 60 – 15 = 45.
\n(\\(\\frac{d^{2}p}{dx^{2}}\\))x = 15<\/sub> = 12(60 – 15) (15 – 30) < 0
\n\u21d2 p is maximum
\n\u2234 Required numbers are 15, 45.<\/p>\n

Question 5.
\nFrom a rectangular sheet of dimensions 30 cm x 80 cm four equal squares of side x cm are removed at the comers and the sides are then turned up so as to form an open rectangular box. Find the value of x, so that the volume of the box is the greatest?
\nSolution:
\nLength of the box = 80 – 2x = l
\nBreadth of the box = 30 – 2x = b
\n\"Inter
\nHeight of the box = x = h
\nVolume = lbh = (80 – 2x) (30 – 2x). x
\n= x (2400 – 220 x + 4x\u00b2)
\nf(x) = 4x\u00b3 – 220 x\u00b2 + 2400x
\nf'(x) = 12x\u00b2 – 440x + 2400
\n= 4 [3x\u00b2 – 110 x + 600]
\nf'(x) = 0
\n\u21d2 3x\u00b2 – 110 x +600 = 0
\n\"Inter
\nf(x) is maximum when x = \\(\\frac{20}{3}\\)
\nVolume of the box is maximum when x = \\(\\frac{20}{3}\\) cm<\/p>\n

Question 6.
\nA window is in the shape of a rectangle surmounted by a semi-circle. If the peri-meter of the window is 20 feet, find the maximum area.
\nSolution:
\nLet length of the rectangle be 2x and breadth be y so that radius of the semi-circle is x.
\n\"Inter
\nPerimeter = 2x + 2y + \u03c0. x = 20
\n2y = 20 – 2x – \u03c0x.
\ny = 10 – x – \\(\\frac{\\pi}{2}\\). x
\nArea = 2xy + \\(\\frac{\\pi}{2}\\). x\u00b2
\n= 2x(10 – x – \\(\\frac{\\pix}{2}\\)) + \\(\\frac{\\pi}{2}\\)x\u00b2
\n= 20x – 2x\u00b2 – \u03c0x\u00b2 + \\(\\frac{\\pi}{2}\\) x\u00b2
\nf(x) = 20x – 2x\u00b2 – \\(\\frac{\\pi}{2}\\) x\u00b2
\nf'(x) = 0 \u21d2 20 – 4x – \u03c0x = 0
\n(\u03c0 + 4)x = 20
\nx = \\(\\frac{20}{\\pi+4}\\)
\nf”(x) = -4 – \u03c0 < 0
\nf(x) is a maximum when x = \\(\\frac{20}{\\pi+4}\\)
\n\"Inter<\/p>\n

Question 7.
\nIf the curved surface of right circular a cylinder inscribed in a sphere of radius r is maximum, show that the height of the cylinder is \u221a2r.
\nSolution:
\nSuppose r is the radius and h be the height of the cylinder.
\n\"Inter
\nFrom \u2206OAB, OA\u00b2 + AB\u00b2 = OB\u00b2
\nr\u00b2 + \\(\\frac{h^{2}}{4}\\) = R\u00b2 ; r\u00b2 = R\u00b2 – \\(\\frac{h^{2}}{4}\\)
\nCurved surface area = 2\u03c0rh
\n\"Inter
\nf(h) is greatest when h = \u221a2 R
\ni.e., Height of the cylinder = \u221a2 R.<\/p>\n

\"Inter<\/p>\n

Question 8.
\nA wire of length l is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of the pieces of the wire respe-ctively so that the sum of the areas is the least?
\nSolution:
\nSuppose x is the side of the square and r is the radius of the circle.
\n\"Inter
\nGiven 4x + 2\u03c0r = l
\n4x = l – 2\u03c0r
\nr = \\(\\frac{1-2 \\pi r}{4}\\)
\nSum of the area = x\u00b2 + \u03c0r\u00b2
\n\"Inter
\n\u2234 f(r) is least when r = \\(\\frac{l}{ 2(\\pi+4)}\\)
\nSum of the area is least when the wire is cut into pieces of length \\(\\frac{\\pi l}{\\pi+4}\\) and \\(\\frac{4l}{\\pi+4}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h) I. Question 1. Find the points of local extrema (if any) and local extrema of the … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22164"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22164"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22164\/revisions"}],"predecessor-version":[{"id":22165,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22164\/revisions\/22165"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22164"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22164"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}