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29) The maximum value of $[ x ( x-1)+ 1 ] ^{1/3 } , 0\leq x \leq 1$
$(A) \left ( \frac{1}{3} \right ) ^{1/3}\: \: (B) 1 /2\: \: (C) 1\: \: (D) 0$

Given function is Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0 Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1 option c is correct

28) For all real values of x, the minimum value of
$\frac{1- x + x^2 }{1+ x +x^2}$
is
(A) 0    (B)   1   (C) 3 (D) 1/3

Given function is              Hence, x = 1 and x = -1 are the critical points Now, Hence, x = 1 is the point of minima and the minimum value is  Hence, x = -1 is the point of maxima Hence,  the minimum value of   is    Hence, (D) is the correct answer

27) The point on the curve $x^2 = 2y$ which is nearest to the point (0, 5) is

$(A) (2 \sqrt 2,4) \: \: (B) (2 \sqrt 2,0)\: \: (C) (0, 0)\: \: (D) (2, 2)$

Given curve is  Let the points on curve be   Distance between two points is given by    Hence, x = 0 is the point of maxima Hence, the point  is the point of minima Hence, the point  is the  point on the curve  which is nearest to the point (0, 5) Hence, the correct answer is (A)

26) Show that semi-vertical angle of the right circular cone of given surface area and
maximum volume is $\sin ^{-1} (1/3)$

Let r, l, and h are the radius, slant height and height  of cone respectively Now, Now, we know that The surface area of the cone (A) =  Now,  Volume of cone(V) =   On differentiate it w.r.t to a and after that we will get  Now, at   Hence, we can say that  is the point if maxima  Hence proved

25) Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt 2$

Let a be the semi-vertical angle of cone Let r , h , l are the radius , height , slent height  of cone Now, we know that Volume of cone (V) =  Now, Now, Now, at  Therefore,   is the point of maxima Hence proved

24) Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt 2$ time the radius of the base.

Volume of cone(V)   curved surface area(A) =  Now , we can clearly varify that when  Hence,  is the point of minima Hence proved that  the right circular cone of least curved surface and given volume has an altitude equal to  time the radius of the base

23) Prove that the volume of the largest cone that can be inscribed in a sphere of radius r is 8/27  of the volume of the sphere.

Volume of cone (V) =  Volume of sphere with radius r =  By pythagoras  theorem in  we ca say that V =  Now, Hence, point  is the point of maxima Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is   Volume =  Hence proved

22) A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Area of the square (A) =  Area of the circle(S) =  Given the length of wire  = 28 m Let the length of one of the piece is x m  Then the length of the other piece is (28 - x) m Now,   and  Area of the combined circle and square  = A + S                                                            Now, Hence,   is the point of minima Other length is = 28 - x                        ...

21) Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Let r be the radius of base and h be the height of the cylinder The volume of the cube (V) =  It is given that the volume of cylinder = 100  Surface area of cube(A) =               Hence,  is the critical point Hence,  is the point of minima Hence,  and  are  the dimensions of the can which has the minimum surface area

20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Let r be the radius of the base of cylinder and h be the height of the cylinder we know that the surface area of the cylinder  Volume of cylinder   Hence,  is the critical point Now, Hence,   is the point of maxima  Hence,  the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base

19) Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively and the radius of the circle is r Now, by Pythagoras theorem a = 2r Now, area of reactangle(A) = l b Now, Hence,   is the point of maxima Since, l = b we can say that the given rectangle is a square Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area

18) A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?

It is given that the sides of the rectangle are 45 cm and 24 cm Let assume  the side of the square to be cut off is x cm Then, Volume of cube                 But  x cannot be equal to 18 because then side (24 - 2x) become negative which is not possible so we neglect value x= 18 Hence, x = 5 is the critical value Now, Hence, x  = 5 is the point of maxima Hence,  the side of the...

17. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

It is given that the side of the square is  18 cm Let assume that  the length of the side of the square to be cut off is x cm So, by this, we can say that the breath of cube is (18-2x) cm and height is x cm Then, Volume of cube =     But the value of x can not be 9 because then the value of breath become 0 so we neglect value x = 9 Hence, x = 3 is the  critical point Now, Hence, x = 3...

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

let x an d y are positive two numbers  It is given that  x + y = 16 , y = 16 - x  and   is minimum Now,  Hence, x = 8 is the only critical point  Now, Hence, x = 8 is the point of minima  y = 16 - x     = 16 - 8 = 8 Hence, values of x and y are 8  and 8 respectively

15) Find two positive numbers x and y such that their sum is 35 and the product $x^2 y^5$
is a maximum.

It is given that x + y = 35 , x = 35 - y and  is maximum Therefore, Now, Now, Hence, y = 35 is the point of minima Hence, y= 0 is neither point of maxima or minima Hence, y = 25 is the point of maxima x = 35 - y    = 35 - 25 = 10  Hence, the value of x and y are 10 and 25 respectively

14) Find two positive numbers x and y such that x + y = 60 and $xy^3$ is maximum.

It is given that x + y = 60 , x = 60 -y and  is maximum let  Now, Now, hence, 0 is niether point of minima or maxima Hence, y = 45 is point of  maxima x = 60 - y   = 60 - 45 = 15 Hence, values of x and y are 15 and 45 respectively

14. Find two numbers whose sum is 24 and whose product is as large as possible.

Let x and y are two numbers It is given that x + y = 24 , y = 24 - x and product of xy is maximum  let  Hence,  x = 12 is the  only critical value Now, at x= 12   Hence, x = 12 is the point of maxima Noe, y = 24 - x             = 24 - 12 = 12 Hence, the value of x and y are 12 and 12 respectively

12. Find the maximum and minimum values of  $x + \sin 2x \: \:on \: \: [ 0 , 2 \pi ]$

Given function is So, values of x are   These are the critical points of the function  Now, we need to find the value of the function  at    and at the end points of given  range i.e. at x = 0 and  Hence, at    function  attains its maximum value and value is  in the given range of  and at x= 0 function  attains its minimum value and value is 0

11. It is given that at x = 1, the function $x ^4 - 62x^2 + ax + 9$attains its maximum value,
on the interval [0, 2]. Find the value of a.

Given function is  Function  attains maximum value at x = 1 then x must one of the critical point of the given function that means Now, Hence, the value of a is 120

10. Find the maximum value of $2 x^3 - 24 x + 107$ in the interval [1, 3]. Find the
the maximum value of the same function in [–3, –1].

Given function is     we neglect the value x =- 2 because   Hence, x = 2  is the only  critical value  of function  Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3             Hence, maximum value of function    occurs at x = 3 and vale is 89  when  Now, when  we neglect the value x = 2  Hence, x = -2  is the only  critical value ...
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