MAXIMA AND MINIMA - 2

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Curtis Dawson
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1 MAXIMA AND MINIMA - GREATEST AND LEAST VALUES Definition: Let f be a function defined on a set A and l f( A ). Then l is said to be (i) the maimum value or the greatest value of f in A if f( ) l A. (ii) the minimum value or the least value of f in A if f( ) l A LOCAL MAXIMUM AND LOCAL MINIMUM VALUES Let f be a function defined in a nbd of a point a then f is said to have (i)a local maimum (value) or a relative maimum at a if a δ > such that f( ) < f( a) ( a δ, a) ( a, a+δ ). In this case a is called a point of local maimum of f and f( a ) is its local maimum value. (ii) a local minimum (value) or relative minimum at a if a δ > such that f( ) > f( a) ( a δ, a) ( a, a+δ ). In this case a is called a point of local minimum and f( a ) is its local minimum value.

2 THEOREM: Let f be a differentiable function in a nbd of a point a. The necessary condition for f to have local maimum or local minimum at a is f ( a ) =. FIRST DERIVATIVE TEST Let f be a differentiable function in a nbd of a point a and f ( a ) =. Then (i) f( ) has a relative maimum at =a if a δ > such that ( a δ, a) f ( ) > and ( a, a+δ) f ( ) <. (ii) f( ) has a relative minimum at = a if a δ > such that ( a δ, a) f ( ) < and ( a, a+δ) f ( ) >. (iii) f( ) has neither a relative maimum nor a minimum at =a if f ( ) has the same sign for all ( a δ, a) ( a, a +δ). SECOND DERIVATIVE TEST Let f( ) be a differentiable function in a nbd of a point a and let f ( a ) eist. (i) If f ( a ) = and f ( a ) < then f( ) has a relative maimum at f( ) and the maimum value at a is f( a ). (ii) If f ( a) = and f ( a ) > then f( ) has a relative minimum at f( ) and the minimum value at a is f( a ).

3 ABSOLUTE MAXIMA AND ABSOLUTE MINIMA Let f be a function defined on [a, b]. Then (i) Absolute maimum of f on [a, b] = Ma.{ f( a), f( b ) and all relative maimum values of f in( ab, )}. (ii) Absolute minimum of f on [a, b] = Min.{ f( a), f( b) and all relative minimum values of f in ( ab, )}. Note : 1) maimum and minimum values are called etrimities. ) If f 1 (a) =, then f is said to be stationary at a and f(a) is called the stationary value of f. and (a, f(a) ) is called a stationary point of f. Eercise 1. Find the point at which the local maima or local minima (if any) are attained for the following function. Also find the local maimum or local minimum values as the case man be. i) f() = - Sol. f() = - 1 f () = and f 11 () = 6 For maimum or minimum 1 f () = = 1= =± 1 " f (1) = 6(1) = 6> f() has minimum at = 1and that minimum value is f(1) = 1 (1) = f ( 1) = 6( 1) = 6 < f() has maimum value at = -1 And that maimum value is f( 1) = ( 1) ( 1) = -1 + =

4 ii) f() = Sol. f() = For maimum or minimum f() = = 4+ = ( 1)( ) = = 1or Now f () 1 = 6() 1 1= 6< f( ) has maimum value at = 1 Ma. value is () () () f () = and f 1 = = = 19 f ( ) = 6( ) 1 = 18 1 = 6 > f( ) has minimum value at = " f () = 6 1 Min. value is f ( ) = = = 15 iii) f( ) = ( 1)( + ) ans. f has min. value at = and that min = -4 F has ma. value at = and that is iv) f( ) = + ( > ) Sol: f( ) = + ( > ) 1 4 = = f ( ) and f ( ) For ma. or min. f ( ) = 1 = 4 = =±

5 4 1 f = = > Since > ( ) ( ) f( ) has min. value at = Min. value is ( ) v) f( ) = 1 + ans: ma f() = 1/ f = + = 1+ 1= vi) f( ) = 1 ( < < 1). vii) f( ) = ( 1) ( + 1) critical values at =, min value at =-1 and that min =, ma at 1 = 5 and ma = 456/15. Prove that the following functions do not have maima or minima. i) ( ) f = e Sol: ( ) ( ) f = e and f = e ii) f( ) For maima or minima f ( ) = e = which is not true i.e., there is no value of satisfying f 1 () =. Hence it has no maima or minima. = log = 1 = 1 o for all. Sol: f ( ) and f ( ) f ( ) f ( ) has no maima or minima.

6 iii) ( ) f 1 = Sol: f ( ) = o for all. f ( ) It has no maimum or minimum values.. Find the absolute maimum value and the absolute minimum value of the following functions in the given intervals.? i) f( ) = in [, ] Sol: ( ) ( ) f = and f" = 6 And > for all vaules of. hence f is an increasing function. Therefore it greatest value is f() and least value is f(-) f ( ) ( ) = = 8 And ( ) ii) f( ) = ( 1) + in [,1] f = = 8 Sol: f( ) = ( 1) + in [,1] f ( ) = ( 1) f ( ) = =1. Now f ( ) = ( 1) + = 16+ = 19 f () 1 = ( 1 1) + = + = Ma. value = 19 Min.value = iii) f( ) = sin+ cos in [, π ] 1 Sol: f( ) = sin+ cos in [, π] ( ) Now f ( ) = cos sin = Now f ( ) = sino+ cos= + 1= 1 f = cos sin π = 4

7 π π π f = sin + cos = = + = = f ( π ) = sinπ+ cosπ= 1= 1 π Abslolute ma = ma of { f(),f( ),f(π)} = ma of 4 { 1,, 1} = Absolute min = min of { f(),f( 4 π ),f(π)} = -1. iv) f ( ) = in[ 1,1] ANS: Minimum value is 11 Maimum value is 1. v) f( ) = + sin in [,π ] ANS: Minimum value = II Maimum value is = π 1. Find two positive numbers whose sum is 1 and the sum of the squares is minimum. Sol: let and y be the given numbers + y = 1 y = 1 Let f( ) = + y f( ) = + ( 1 ) = = f " = 4 f ( ) = 4 4and ( ) For ma or min, f ( ) = 4 4= = = 6 4

8 f" ( ) = 4> f( ) has minimum when = 6 y = 1 = 1 6 = 6 The numbers are 6, 6.. Find two positive numbers and y such that + y = 6 and y is maimum. Sol: let and y be the given numbers + y = 6 y = 6 ---(1) Let f( ) = y = ( 6 ) ( ) ( ) ( ) f ' = ( ) ( ) = ( 6 ) [ 6 ] = + ( 6 ) ( 6 4) 4( 6 ) ( 15 ) = = f" = 4 ( 6 ) ( 1) + ( 15 ) ( 6 )( 1) = 4( 6 )[ 6+ + ] = 4( 6 )( 9) = 1( 6 )( ) For maimum or minimum f' = ( ) ( ) = = 6 or = 15; cannot be 6 = 15 y = 6 15 = 45 f "( 15) = 1( 6 15)( 15 ) < f has maimum when =15

9 y =45. Required numbers are 15, 45.. Find the shortest distance from ( 6,) to + y =. Sol: Equation of the line is + y = =. let A( 6,) y let P(,y) be any point on the line P(, y) A( 6, ) + y = ( ) AP = y ( 6) ( ) = + + = = Let ( ) f 5 45 = + f ( ) = 1 and f" = 1 For ma or min f ( ) = o 1 = = f" ( ) = 1> f() is minimum when = AP = 45 AP = 45 = 5 Shortest distance = 5 units.

10 4. Find the shortest distance from ( 6,) to Sol: let P(,y) be any point on the line y + 16= y = + 16 A( 6,) is the given point. y + 16=. ( ) ( ) AP = y = = = Let ( ) f 1 5 = + + f ( ) = and f"=4 For ma or min, f ( ) = 4+ 1= 4 = 1 1 = = 4 f" ( ) = 4> f( ) is minimum when = X= - y = + 16= 9+ 16= 5; y = 5 Shortest distance AP = ( + 6) + 5 = 9+ 5 = 4

11 5. Find the dimensions of the right circular cylinder of greatest volume, that can be inscribed in a sphere of radius a. Sol: Suppose R is the radius and H is the height of the cylinder. Then OA = H/. From Δ OAB, OB = OA + AB a R H = R + 4 H = a Volume of the cylinder V = 4 V =πh a Let ( ) H 4 f H =π a H 4 ( ) H H f H =π a 4 π R H, here H is the variable. And f" ( H) 6H =π 4 For ma or min f 1 = H π a = 4 H H a = ah 4 4 =

12 4a a H = H= 6H f" ( H) =π < 4 f( H ) is maimum when H = a H a a R = a = a = 4 R = a Dimensions of greatest cylinder are Base of the cylinder = a Height of the cylinder = a 6. For what values of >, the ratio of ln to is greatest Sol: Let f ( ) n = l 1 l n.1 1 l n = = f ( ) 1 l n f = = For ma or min ( ) 1- log = = e ( ) f" 1 ( 1 l n) = 4 = ( ln)

13 1 f "( e) = = < e e f( ) is greatest when = e i.e., l n is greatest when = e. III. 1. From a rectangular sheet of dimensions cm 8cm four equal squares of side cm are removed at the corners and the sides are then tuned up so as to form an open rectangular bo. What is the value o, so that the volumes of the bo is the greatest? Sol: length of the sheet = 8, breadth =. Let the Side of the square = Length of the bo = 8 =l Breadth of the bo = = b 8 Height of the bo = = h Volume = l = ( )( ) bh 8. ( ) = f ( ) = f ( ) = and f " = 4( 6 11) = For ma or min f ( ) = =

14 11 ± 11 7 = 6 11 ± = = or = or If =, b= = ( ) = < = = f '( ) = 4 44 =,f ' = When ( ) = = 8 < f() is maimum when = Volume of the bo is maimum when = cm.. A window is in the shape of a rectangle surmounted by a semi-circle. If the perimeter of the window be feet, find the maimum area. Sol: Let the length of the rectangle be and breadth be y so that radius of the semicircle is. Perimeter = + y +π. = y = π π y= 1.

15 π π π Area = y +. = 1 + π = π + Let f() π = π + f ' = 4 π+ π and f" =-4- π+ π= -4-π for ma or min for ma or min f ( ) = 4 π = ( π+ 4) = = π+4 f' ( ) = 4 π< f( ) has a maimum when = π+4 π π y= 1 = 1 π+ 4 π+ 4 1π+ 4 1π = π+ 4 = π+4 Maimum area π = y +. 4 π 4 =. + π+ 4 π+ 4 π+ 4 ( ) ( ) ( π+ ) ( ) 8 + π 4 = = π+ 4 π+ 4 = sq.feet. π+ 4

16 . Show that when curved surface area of a cylinder inscribed in a sphere of radius R is a maimum, then the height of the cylinder is R. Sol: let r be the radius and h be the height of the cylinder. O H A r R B From Δ OAB, OA + AB = OB h h r + = R ; r = R 4 4 Curved surface area = π rh h = π R.h 4 =πh 4R h Let f ( h) =πh 4R h 1 f h =π h. h + 4R h.1 4R h ( ) ( ) ( ) h + 4R h π R h =π. = 4R h 4R h For ma or min f ( h) = R h = ( ) π R h = 4R h h = R h = R

17 4R h ( h) + ( R h ) d 4R h f'h = π dh 4R h And ( ) 4π h+ = < whenh R = 4R f( h ) is greatest when h = R i.e., Height of the cylinder = R 4. A 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 wire so that the sum of the areas is least Sol: Suppose is the side of the square and r is the radius of the circle. Given 4 + π r =l 4 = l πr πr = l 4 Sum of the areas = +π r Let f( r) ( l πr) = +πr 16 ( l π r) Then f '( r) = ( π) + πr 16 π and f "( r) = + π For ma/min f ( r ) = ( ) ( l π ) ( ) r f r = π + π r = 16 π f r = π + π r = 4 ( ) ( l ) π π r = π r 4 ( l )

18 l r = π+ 4 ( ) l πr 1 πl = = l 4 4 π+ 4 π+ l rl πl 4l = = 4π+ 4 4 π+ 4 ( ) = π+ l 4 4l 4 = π+ 4 And f () > ( r) f is least when l r = π+ 4 ( ) Sum of the area is least when the wire is cut into pieces of length. πl 4l and π+ 4 π+ 4 Equation 1 1. Find the intervals in which the following function are strictly increasing or strictly decreasing. 6 9 Sol. Let f() = 6 9 f () = 9 f() is increasing if f () > 9 9 > + 9 < < 9 f() is increasing if, f() is decreasing if f () < > > 9 f() is decreasing of,.. Find the intervals in which the function f() = sin 4 + cos 4 π, is increasing and decreasing.

19 Sol. f() = sin 4 + cos 4 f() = (sin ) + (cos ) = (sin + cos ) sin cos 1 = 1 sin 1 f() = sin cos() = sin cos = 4sin4 π Let < < 4 f() is decreasing if f () < sin < sin > π, 4 f() is increasing if f () > sin > sin < π π, 4 Find the points of local etrema (if any) and local etrema of the following functions each of whose domain is shown against the function.. f() = sin, [, 4π). Sol. Given f() = sin f() = cos f() = sin For ma or min, f() = cos = π π 5π 7π =,,, π π i) f = sin = 1<

20 π f() = sin = 1 Point of local maimum Local maimum = 1 π π ii) f = sin = 1> π = π f() = sin = 1 Point of local minimum = π Local minimum = 1 5π 5π iii) f = sin = 1< 5π f() = sin = 1 Point of local maimum = 5 π Local maimum = 1 7π 7π iv) f = sin = 1> 7π f() = sin = 1 Point of local minimum = 7 π Local minimum = 1 4. f() = ( 1 ) (, 1) 1 Sol. f() = ( 1) = = = 1 1 For ma. or min. f () =

21 = = = 1 1 f = 1 = = 5. Use the first derivative test to find local etrema of f() = on R. Sol. f() = f () = 6 f () = for maimum or minimum f () = 6 = = f () = > Point of local minimum =. Local minimum = Find local maimum or local minimum of f() = sin defined on [ π/, π/]. Sol. f() = sin f () = cos 1 f () = 4 sin π π Thus the starting point are =, at π π π 4 =,f = 4sin = > π π π f = sin π π π 4 at = f = 4sin = < π π π π f =+ sin + = +

22 π Local minimum = π Local maimum = + 7. Find the maimum profit that a company can make, if the profit function is given by P() = Sol. P() = dp() 7 6 d = For maima or minima, dp d = 7 6 = = dp 6 d = < The profit f() is maimum for = The maimum profit will be P() = () 18(4) = 1 8. The profit function P() of a company selling items per day is given by P() = (15 ) 1. Find the number of items that the company should manufacture to get maimum profit. Also find the maimum profit. Sol. Given that the profit function P() = (15 ) 1 For maimum or minimum dp d = (15 (1) ( 1) = 15 = = 75 dp Now d = < The profit P() is maimum for = 75 The company should sell 75 terms a day The maima profit will be P(75) = 465.

23 9. Find the absolute maimum and absolute minimum of f() = on [ 8, ]. Sol. f() = f () = For maimum or minimum, f () = = = 4( + 7) 1( + 7) = ( + y)(4 1) = 1 = 7 or 4 f ( 8) = 8( 8) + 81( 8) 4( 8) 8 = 8(51) + 81(64) = = = 1416 f () = 8() + 81() 4() 8 = = f = = = = = f ( 7) = 146 Absolute maimum = Absolute minimum = Find the maimum area of the rectangle that can be formed with fied perimeter. Sol. Let and y denote the length and the breadth of a rectangle respectively. Given that the perimeter of the rectangle is. i.e., ( + y) = i.e., + y = 1 (1) Let A denote the area of rectangle. Then

24 A = y () Which is to be minimized, equation (1) can be epressed as y = 1 () from () and (), we have A = (1 ) A = 1 (4) Differentiating (4) w.r.t. we get da 1 d = (5) The stationary point is a root of 1 = = 5 is the stationary point. Differentiating (5) w.r.t., we get da d = which is negative. Therefore by second derivative test the area A is maimized at = 5 and hence y = 1 5 = 5, and the maimum area is A = 5(5) = Find the point on the graph y = which is the nearest to the point (4, ). Sol. Y P(,y) P(,y) P(,y) O A(4,) X Let P(, y) be any point on y = and A(4, ). We have to find P such that PA is minimum. Suppose PA = D. The quantity to be minimized is D. D = ( 4) + (y ) (1) P(, y) lies on the curve, therefore y = () from (1) and (), we have

25 D = ( 4) + D = ( ) () Differentiating () w.r.t., we get dd 7 1 = d Now dd d = Gives = 7/. Thus 7/ is a stationary point of the function D. We apply the first derivative test to verify whether D is minimum at = 7/ dd 1 1 = d = and it is negative dd 1 1 = d = and it is positive dd changes sign from negative to positive. Therefore, D is minimum at = d 7/. Substituting = 7/ in () we have y = 7/. y =± 7 Thus the points 7 7, and 7 7, are nearest to A(4, ). 1. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Sol. Let O be the center of the circular base of the cone and its height be h. Let r be the radius of the circular base of the cone. Then AO = h, OC = r Let a cylinder with radius (OE) be inscribed in the given cone. Let its height be u.

26 A P R Q B D O E C i.e. RO = QE = PD = u Now the triangles AOC and QEC are similar. Therefore, QE = EC OA OC i.e., u r = h r h(r ) u = r Let S denote the curved surface area of the chosen cylinder. Then S = π u As the cone is fied one, the values of r and h are constants. Thus S is function of only. ds d S 4πh Now, = πh(r )/r and = d d r The stationary point of S is a root of ds d = i.e., π(r ) / r = r i.e., = ds d < for all, therefore ds d = r/ < Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is r/.

27 Additional Problems for Practice 1. Show that the semi-vertical angle of the right circular cone of maimum 1 volume and of given slant height is tan. Sol: let r be the base, h be the height, l be the slant height and α be the semi-vertical angle of the cone. O h α P A N B From r h ΔOAB, sin α =, cos α = l l Volume of the cone is ( l sin )( l cos ) π = α α 1 v = Ar h πl = sin α.cos α Let ( ) πl f α =.sin α.cosα πl f α = (sin α sinα + cosα sinαcosα ( ) ( ) ( ) πl = sin α cos α sin α πl And f ( α ) = {sin α( 4cos αsin α sin αcos α) For ma or min f ( α ) = ( cos sin cos ) + α α α ( ) πl sin α cos α sin α = sin α= or cos α sin α=

28 sin α= cos α tan α= 1 ( ( )) tan α= α= tan When 1 α= tan, cos α= sin α= πl f ( α ) = 6sin αcos α / < f ( α ) is maimum, when 1 α= tan The volume of the cure is maimum when semi-vertical angle is 1 tan. 6. Assume that the petrol burnt (per hour) in driving a motor boat varies as cube of its velocity. Show that the most economic speed. When going against a current of 6k per hour is 9 km. per hour. Ans: most economical speed is 9 kmph. 7. Let A(,a ), B(,b ) be two fied points P(, ) a variable point. Show that when acute angle APB is maimum, = ab. 8. Show that in the area of a rectangle inscribed a circle is maimum when it is a square.? Sol: Let r be the radius of the circle be the length and y be the breadth of the rectangle. Then diagonal of rectangle = diameter of the circle =r. D C r A r O B From Δ ABC, + y = 4r y = 4r Area A = y = 4r Let f ( ) = 4r

29 1 f =. + 4r 1 4r ( ) ( ) ( ) + 4r r = = 4r 4r For ma or min f ( ) = ( ) r = 4r.r = = r = r 4r. r f = ( ) ( ) ( ) ( 4r ) ( ) 4r When = r, we have r = 4 4r 4 f ( ) = = < 4r 4r f is ma when = r y = 4r = 4r r = r y = r Therefore = y = r Hence the rectangle is a square. 9. Find the rectangle of maimum perimeter that can be inscribed in a circle. Sol: let be the length, y be the breadth of the rectangle, r be the radius of the circle. Then diagonal = r.

30 D C r A r O B From Δ ABC, + y = 4r ; y = 4r Perimeter = ( + y) = ( + 4r ) Let f ( ) = ( + 4r ) f = 1 ( ) 4r For ma or min f ( ) = 1 = 4r 4r = 1 = 4r = 4r = r ( ) 4r.1 4r ( ) f ( ) = 4r ( ) 4r = / ( ) ( 4r ) 4r = + 4r /

31 ( ) 4r f' r = < ( r ) 4r Therefore f is ma when = r. y = 4r = 4r r = r y = r i.e., = y f() is maimum when = y i.e., when the rectangle becomes a square. 1. Show that f( ) = sin( )( 1+ cos) has a maimum value at =π / Sol: Given f( ) = sin( 1+ cos) f ( ) = sin ( sin ) + ( 1+ cos ).cos = cos + cos sin = cos + cos 1+ cos = cos + cos 1 For ma or min f ( ) = cos + cos 1 = ( cos + 1)( cos 1) = 1 cos = 1 or cos = Now f ( ) = 4cos.sin sin = sin ( 4cos + 1) when 1 cos =, sin =

32 f ( ) = cos ( + 1) < f( ) is maimum when i.e., π = 1 cos = m n 11. Show that f( ) = sim.cos has maimum value of tan,( mn ) Sol: Given ( ) m n f = sim.cos m n 1 ( ) ( ) f = sin.cos sin n m 1 + cos.m sin ( ) m 1 n 1 = sin.cos. m cos n sin For ma or min f ( ) = ( ) m1 n1 sin.cos m cos n sin =.cos 1 m = >. n m m cos.n sin = tan = n = tan 1 m n m ( ) 1 n 1 ( ) f = sin.cos m cos sin n sin cos ( ) d ( m 1 n m cos n sin. sin.cos 1 ) + m 1 n 1 d ( ) = sin.cos m + n ( ) d sin + m cos n sin n d ( sin m 1.cos n 1 )

33 when 1 m = tan,mcos nsin = n m1 n1 ( ) ( ) f = m + n sin.sin.cos < f( ) is maimum at = tan 1 m n 1. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 of the volume of the sphere. 7 Sol: Let R be the base radius, be the distance of the centre of the sphere from the base and V be the volume of the cone. Height h of the cone R + R O A R r B From Δ OAB, r = R 1 π π = π = ( + )( ) = ( R + R R ) V r h R R ( ) dv π = R R d dv π = d ( R 6) For maimum volume, dv d = R R = ( R+ )( R ) =

34 R =, R d V π =, = R π < d When ( ) V is maimum where R = Ma. volume π R R = R R+ 9 R 4R 8 4 = π.8 = πr = (Volume of the sphere) 7. Find the intervals in which f ( ) = is increasing and the intervals in which f( ) is decreasing.. Show that for all values of a and b, f( ) = + a + a + a + b is increasing. Sol: ( ) f a a a b = f ( ) = + 6a + a ( ) = + a+ a = ( + a) For all values of, the function f is increasing. 4. Show that l n( 1 ) + 1 ( + ) for all. 5. Show that for all.

35 π 6. If show that sin. 7. Find all the local maima and local minima of the function f( ) = Find al he points as local maima and local minima of the function 4 f( ) = Find all the local maima and local minima of the sine function Find the absolute maimum and value of also its absolute minimum value on this interval. Sol: Let ( ) 4 f = ( ) ( ) f = 4 = 1 19 ( ) ( ) f = 1 = on the interval [ ],1. Find 1 1 = or = i.e., = 1 = and = are the two points locate f take the value zero with these two points, we consider the end points and 1 abc of interested. Totally we get only there points. At these three points, we calculate the value of. f ( ) =, f ( 1) = f = = = = 4 4 The maimum value of f(,1 ) 1 = ma f ( ) f,f ( )

36 1 = ma,, = 4 attained at = and = 1 The maimum value of f over (,1 ) 1 = min f ( ),f,f () 1 1 = min,, 4 1 = attained at 4 1 = 11. Find the maimum and minimum values of sin + sin over [,π ] Sol: Let f ( ) = sin+ sin Then f ( ) = cos+ cos f ( ) = cos + cos = cos + cos 1 = ( cos 1)( cos+ 1) = 1 cos = or cos = 1 1 cos in, are = π The solutions of [ π ]

37 π 5π =, The solution of cos = -1 in [, π ] is =π End points of the interval are, π Hence, we have to find the values of the given functions at π 5π,, π, and π π f ( ) =,f =, 5π f ( π ) =,f = and f( π ) =. Maimum value of over [, π ] π 5π = ma. f ( ),f f,f ( π) = ma.,, = Similarly, the minimum value of f over [, π ] = Min.,, = 1. Find the position numbers whose sum is 4 and whose product as large as possible. 1. The sum of the height and diameter of the base of a right circular cylinder in given as units. Find the radius of the base and height of the cylinder so that the volume is maimum. 14. Show that e has the maimum value at = A jet of an enemy is flying along the curve y = +. A soldier is placed at the point (, ). What is the nearest distance between the soldier and the jet?

38 Sol: For each value of, the position of the jet is (, ) the distance between their position and the soldier. ( ) = ( ) + ( + ) f ( ) 4 = + f ( ) = ( ) + 4 f ( ) = + 1 f ( ) = onlyif ( ) + 4 = + = ( )( ) = y +. Let ( ) f be the square of (1, ) (, ) The only point at which f ( ) Since there are no real rots of f () 1 + 1= 14> takes the value zero is = = Hence, at =1 the function has local minimum These are the least value of f f () 1 = ( 1 ) ( 1 ) = + + = 4+ 1= 5 Required minimum distance f ( ) = 5. Observe that there is no maimum value in this problem.

3. Total number of functions from the set A to set B is n. 4. Total number of one-one functions from the set A to set B is n Pm

3. Total number of functions from the set A to set B is n. 4. Total number of one-one functions from the set A to set B is n Pm ASSIGNMENT CLASS XII RELATIONS AND FUNCTIONS Important Formulas If A and B are finite sets containing m and n elements, then Total number of relations from the set A to set B is mn Total number of relations