 Hello everyone, Myself A.S. Falmari, Assistant Professor, Department of Humanities and Sciences, Balchand Institute of Technology, Solapur. In the last video, we have started to study Lagrange's undeterred mind multiplier. In that video, we have seen the method of Lagrange's undeterred mind multiplier and few examples. Now, in this video also, we will take some more examples. Now, the learning outcome is at the end of this session, students will be able to find the stationary values of given functions subjected to some conditions. Now, pause this video and answer this question. What is the volume of a rectangular solid? I hope that all of you have written the answer. Now, the problem is what is the volume of a rectangular solid? If w is a width, h is the height and l is the length of that rectangular solid, then the volume of this rectangular solid is given by v equal to product of the width, height and the length. Now, consider one example. Show that the rectangular solid of maximum value that can be inscribed in a sphere is a cube. Now, here we have to show that the rectangular solid of maximum value that can be inscribed in a sphere is a cube. So, the given function is a rectangular solid. We know that it has the width, height and length. Let us assume that 2x, 2y, 2z be the length, breadth and height of the rectangular solid and the subjected condition here provided is that the sphere. So, we know that it has a radius. Let us assume that r be the radius of this sphere. Now, just we have seen the volume of a rectangular solid which is provided by the product of the length, breadth and the height. Now, these are the length, breadth and the height and it is equal to 8 into x, y, z. Now, we will denote this function by f that is equal to 8 into x, y, z. Now, call it as equation number one. Now, the subjected condition is a sphere. Now, the sphere centered at origin has the equation x square plus y square plus z square is equal to r square. Let me write it in a standard form. Let me denote it as phi of x, y, z equal to x square plus y square plus z square. Taking this r square to the left hand side, we get it as minus r square is equal to 0. Now, call it as equation number two. Now, here as that f and phi are functions in three independent variables, we are having these three Lagrange's equations. Consider the first one. Now, the partial derivative of this function f with respect to x, treating y and z are constant is 8 y z plus lambda times. Now, phi is x square plus y square plus z square minus r square. Its partial derivative with respect to x, treating y and z constant is simple to x. So, here it is 2 into lambda x equal to 0. Now, call it as equation number three. Now, the next equation is dou f by dou y plus lambda times dou phi by dou y equal to 0. Now, it implies differentiating f partially with respect to y. The derivative of y is 1. Now, the 8 x z as it is plus now lambda times partial derivative of phi with respect to y is 2 y. Therefore, it is 2 lambda y equal to 0. Now, call it as equation number four. And the last equation is dou f by dou z plus lambda times dou phi by dou z equal to 0 implies. Now, the partial derivative of f with respect to z is 8 x y plus and lambda times the partial derivative of phi with respect to z is 2 z that is equal to 0. Now, let us call it as equation number five. Now, from the equation number three we can write it as 2 lambda x equal to minus 8 y z. Winning equation can be written like 2 lambda y equal to minus 8 x z 2 lambda z equal to minus 8 x y. Now, actual our aim is to eliminate the parameter lambda and obtain the values of x y z. Multiplying this equation by x we get it as 2 lambda x square equal to minus 8 x y z. Let us call it as equation number six. And to the next equation multiplying by y we get it as 2 lambda y square equal to minus 8 x y z. Call it as next equation number seven. And to the final equation multiplying it by z we get it as 2 lambda z square equal to minus 8 x y z. Now, call it as equation number eight. As in six seven eight right hand sides are equal we can equate the left hand sides. So, after equating left hand sides we get the equation 2 lambda x y equal to 2 lambda y square equal to 2 lambda z square. Now, it implies after removing 2 lambda that is by dividing by 2 lambda we get it as x square equal to y square equal to z square. Now, here x y z are the length, breadth, height therefore, they are always positive. So, that now it will implies that x equal to y equal to z. Now, that whatever may be the quantities x y z height, length, breadth we have considered that are all are equal. It means what that rectangular solid is nothing but a cube. Now, it proves a rectangular solid inscribed in a sphere is a cube. Consider the next problem find the maximum and minimum distances from the origin to the curve 3 x square plus 4 x y plus 6 y square equal to 1 4 0. Solution here we want to extremize the distance and we know that the distance of any point x comma y from the origin is given by d equal to under root of x minus 0 square that is x square plus y minus 0 square is y square. Now, we have to extremize this condition subject to this curve. Now, extremizing this distance is exactly equivalent to extremizing its square. Now, taking square on both sides we get d square equal to x square plus y square. Now, let us denote this function by f which is equal to x square plus y square. Now, let us call it as equation number 1 denote the given condition curve by phi equal to 3 x square plus 4 x into y plus 6 y square minus 1 4 0 after taking 1 4 0 to the left hand side is equal to 0. Now, let us call it as equation number 2. Now, here the functions are in two independent variables. Therefore, we have two Lagrange's equation. First one is dou f by dou x plus lambda times dou phi by dou x equal to 0. It implies differentiating this f partially with respect to x trading y constant we get it as 2 x plus lambda times differentiating phi partially with respect to x trading y constant we get derivative of 3 x square is 6 x plus derivative of 4 x y is 4 y and the derivative of remaining constant is 0 and it is equal to 0. Call it as equation number 3. Consider the next equation dou f by dou y plus lambda times dou phi by dou y equal to 0 which implies now differentiating f partially with respect to y trading x constant we get it as 2 y plus lambda times differentiating phi partially with respect to y trading y constant we get it is a constant its derivative is 0 derivative of 4 x y is 4 x plus derivative of 6 y square is 12 y and the derivative of constant is 0 and it is equal to 0. Let us call it as equation number 4. Now from these two equations we can observe that it is possible to obtain the value of lambda in terms of x y therefore from 3 we get the value of lambda as minus 2 x upon 6 x plus 4 y. Now from the denominator it is possible to take 2 common and it is possible to remove it so we get it as minus x upon 3 x plus 2 y call it as equation number 5 and from the equation number 4 we get the value of lambda as minus 2 y upon 4 x plus 12 y. Again from the denominator we can take 2 common and we can remove these two from the numerator and denominator we get minus y upon 2 x plus 6 y call it as equation number 6. Now from 5 and 6 we can see that both are the values of lambda therefore they must be equal therefore we can write lambda equal to minus x upon 3 x plus 2 y is equal to minus y upon 2 x plus 6 y. Now let us multiply by minus throughout the equation we get minus lambda is equal to x upon 3 x plus 2 y is equal to y upon 2 x plus 6 y. Now for this term multiplying and dividing by x we get minus lambda equal to x square upon 3 x square plus 2 x y and to this quantity multiplying and dividing by y we get y square upon 2 x y plus 6 y square. Now let us adding these two quantities we get minus lambda equal to x square plus y square upon 3 x square plus 6 y square plus 2 x y plus 2 x y is 4 x y. Now we can see that this x square plus y square is nothing but f from equation number 1 and 3 x square plus 6 y square plus 4 x y is the curve and it has the value 1 4 0 from the equation number 2. That is the value of lambda we have obtained minus f upon 1 4 0. From equation number 3 after substituting the value of lambda we get 2 x minus lambda is f upon 1 4 0 into the bracket 6 x plus 4 y is equal to 0. Now multiplying by 1 4 0 2 out this equation we get 2 8 0 into x minus 6 f x minus 4 into f into y is equal to 0. Now again dividing this equation by 2 we get 1 4 0 into x minus 3 f into x minus 2 f into y is equal to 0. Now from the first two equations we can take x common and inside the bracket the left quantity is 1 4 0 minus 3 into f into x minus 2 fy as it is is equal to 0. Call it as next equation number 7. Again substituting the value of lambda in equation number 4 we get 2 y minus f upon 1 4 0 inside the bracket 4 x plus 12 y equal to 0. Again simplifying in the same manner as we have simplified here we get it as minus 2 f into x plus inside the bracket 1 4 0 minus 6 into f into y is equal to 0 equation number 8. From the equation number 7 we can get the value of x as 2 into f into y upon 1 4 0 minus 3 into f. Now in substituting it in equation number 8 we get this minus 2 into f and here the numerator is 2 into f into y we get it as minus 4 f square into y upon 1 4 0 minus 3 into f plus this remaining quantity inside the bracket 1 4 0 minus 6 f into y as it is equal to 0. Now multiplying this equation by 1 4 0 minus 3 into f upon y we get minus 4 into f square plus 1 4 0 minus 6 f into another bracket 1 4 0 minus 3 f equal to 0. Now simplifying bracket we get minus 14 f square minus 1 2 6 0 into f plus 1 4 0 square equal to 0. Again simplifying the above equation we get f square minus 19 into f minus 1 4 double 0 equal to 0. Now it is a quadratic polynomial in f its roots are f equal to 70 comma 20 but the actually f is a d square that is it is a square of the distance so the actual values are root 20 and root 70 root 20 is the minimum one and root 70 is the maximum one therefore we can say that the maximum and minimum distances are root 70 and root 20 respectively.