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G Gautam harsolia
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 

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G Gautam harsolia
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

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G Gautam harsolia
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)

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G Gautam harsolia
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

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G Gautam harsolia
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   

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G Gautam harsolia
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

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G Gautam harsolia
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  

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G Gautam harsolia
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                        ...

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G Gautam harsolia
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

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G Gautam harsolia
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

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G Gautam harsolia
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

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G Gautam harsolia
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...

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G Gautam harsolia
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...

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G Gautam harsolia
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 

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G Gautam harsolia
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  

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G Gautam harsolia
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

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G Gautam harsolia
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  

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G Gautam harsolia
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

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G Gautam harsolia
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

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G Gautam harsolia
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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