Differentiation (Part-38) | S N De

 

In this part, I have discussed about few math problems which I think is essential for  those who have opted for wbcs math (optional).   

1. If $\,\,V=\log(x^3+y^3+z^3-3xyz),\,\,$then prove that $\\(a)\,\, \left(\frac{\partial }{\partial x} +\frac{\partial }{\partial y}+\frac{\partial }{\partial z} \right)V=\frac{3}{x+y+z} \\ (b)\,\,\left(\frac{\partial^2 }{\partial^2 x} +\frac{\partial^2 }{\partial^2 y}+\frac{\partial^2 }{\partial^2 z} \right)V=\frac{-3}{(x+y+z)^2}  $

Solution. (a)  We have, $\,\,V=\log(x^3+y^3+z^3-3xyz) \\ \therefore \frac{\partial V}{\partial x}=\frac{3x^2-3yz}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (1) \\ \text{Similarly,}\,\, \frac{\partial V}{\partial y}=\frac{3y^2-3zx}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (2) \\ \text{and}\,\, \frac{\partial V}{\partial z}=\frac{3z^2-3xy}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (3) \\ \therefore \text{Adding (1),(2),(3), we get}\,\, \left(\frac{\partial }{\partial x} +\frac{\partial }{\partial y}+\frac{\partial }{\partial z} \right)V \\= \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}\\=\frac{3}{x+y+z}$

Sol.(b) Now, we have, $\frac{\partial V}{\partial x}=\frac{3x^2-3yz}{(x^3+y^3+z^3-3xyz)} \\ \therefore \frac{\partial^2V}{\partial x^2}=\frac{(x^3+y^3+z^3-3xyz).6x-(3x^2-3yz)(3x^2-3yz)}{(x^3+y^3+z^3-3xyz)^2} \\= 3. \frac{(2x^4+2xy^3+2xz^3-6x^2yz)-3(x^4-2x^2yz+y^2z^2)}{(x^3+y^3+z^3-3xyz)^2}\\=\frac{-x^4+2xy^3+2xz^3-3y^2z^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (4)\\ \frac{\partial^2V}{\partial y^2}=\frac{(x^3+y^3+z^3-3xyz)(6y)-(3y^2-3zx)(3y^2-3zx)}{(x^3+y^3+z^3-3xyz)^2}\\=3.\frac{(2x^3y+2y^4+2yz^3-6xy^2z)-3(y^4-2xy^2z+z^2x^2)}{(x^3+y^3+z^3-3xyz)^2}\\=3.\frac{-y^4+2x^3y+2yz^3-3z^2x^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (5)$

Similarly, $\,\, \frac{\partial V}{\partial x}=3. \frac{-z^4+2zx^3+2zy^3-3x^2y^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (6)$

On adding (4),(5),(6) we get, $\left(\frac{\partial^2 }{\partial^2 x} +\frac{\partial^2 }{\partial^2 y}+\frac{\partial^2 }{\partial^2 z} \right)V \\=3. \frac{-x^4+2xy^3+2xz^3-3y^2z^2-z^4+2zx^3+2zy^3-3x^2y^2-z^4+2zx^3+2zy^3-3x^2y^2}{(x^3+y^3+z^3-3xyz)^2}\\=-3.\frac{x^4+y^4+z^4-2xy^3-2xz^3+3y^2 z^2-2x^3y-2yz^3+3z^2x^2-2zx^3-2zy^3+3x^2y^2}{(x^3+y^3+z^3-3xyz)^2}\\ = -3.\frac{(x^2+y^2+z^2-xy-yz-zx)^2}{(x+y+z)^2(x^2+y^2+z^2-xy-yz-zx)^2}\\= \frac{-3}{(x+y+z)^2}$

2. If $\,\,r^2=x^2+y^2+z^2,\,\, V=r^3,\,\,$then prove that $\,\,(i)\left(\frac{\partial^2 }{\partial^2 x} +\frac{\partial^2 }{\partial^2 y}+\frac{\partial^2 }{\partial^2 z} \right)V=12r \\ \text{and},\,\, (ii)\,\frac{1}{zx}. \frac{\partial^2V}{\partial z \partial x}+\frac{1}{yz}. \frac{\partial^2V}{\partial y \partial z}+\frac{1}{xy}. \frac{\partial^2V}{\partial x \partial y}=\frac{9}{r}$

Solution.  Since, $\,\,r^2=x^2+y^2+z^2 \\ \therefore 2r. \frac{\partial r}{\partial x}=2x \Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}; \\ \text{Similarly,}\,\, \frac{\partial r}{\partial y}=\frac{y}{r}; \\ \text{and}\,\, \frac{\partial r}{\partial z}=\frac{z}{r} \\  \therefore V=r^3 \\ \Rightarrow \frac{\partial V}{\partial x}=3r^2 . \\ \frac{\partial r}{\partial x}=3r^2. \frac{x}{r}=3rx; \\ \therefore \frac{\partial^2V}{\partial x^2}=\frac{\partial}{\partial x}(3rx)=3r+ 3x.\frac{\partial r}{\partial x} \\=3r+3x.\frac{x}{r} \\ =3. \frac{r^2+x^2}{r}\\  =\frac{3}{r}(r^2+x^2)\cdots \cdots (1)\\ \therefore \frac{\partial^2V}{\partial y^2}= \frac{3}{r}(r^2+y^2)\cdots \cdots (2), \\ \text{and} \,\, \frac{\partial^2V}{\partial z^2}=\frac{3}{r}(r^2+z^2)\cdots \cdots (3)$

Now, $\,\, \frac{\partial^2V}{\partial y \partial z}=\frac{\partial }{\partial y}(\frac{\partial V}{\partial z})=\frac{\partial }{\partial y }(3rz)=3.\frac{\partial r}{\partial y}.z =3. \frac{y}{r}.z \\ \text{or,}\,\, \frac{1}{yz}. \frac{\partial^2V}{\partial y \partial z}=\frac{3}{r} \cdots \cdots (4) \\ \text{Similarly,}\,\, \frac{1}{zx}. \frac{\partial^2V}{\partial z \partial x}=\frac{3}{r}\cdots \cdots (5);\,\,\, s \cdots (4) \\ \text{and,}\,\, \frac{1}{xy}. \frac{\partial^2V}{\partial x \partial y}=\frac{3}{r}\cdots \cdots (6)\\$

So, Adding (4),(5)and(6) we get, $\frac{1}{yz}.\frac{\partial^2V}{\partial y \partial z}+\frac{1}{zx}.\frac{\partial^2V}{\partial z \partial x}+\frac{1}{xy}.\frac{\partial^2V}{\partial x \partial y} \\=\frac{3}{r}+\frac{3}{r}+\frac{3}{r} \\=\frac{9}{r}$

Read more : TOP BOOKS FOR WBCS MAINS EXAMINATION

 3. If $\,\,u=f(r),\,\,\text{where,}\,\, r=\sqrt{x^2+y^2+z^2},  \\ \text{prove that}\,\,\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial x^2}=f''(r)+\frac{2}{r}f'(r).$ 

In particular, prove that if $\,\,f(r)=r^m$,   then the left side $=m(m+1)r^{m-2}$

Solution. $\,\,\text{Since,}\,\, u=f(r) \Rightarrow \frac{\partial u}{\partial x}=f'(r).\frac{\partial r}{\partial x}=f'(r). \frac{x}{r} \\ [\text{Since,}\, r^2=x^2+y^2+z^2 \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}]\\ \therefore \frac{\partial^2u}{\partial x^2}=f'(r). \frac{r.1-x. \frac{\partial r}{\partial x}}{r^2}+\frac{x}{r}.f''(r).\frac{\partial r}{\partial x} \\= \frac{f'(r)}{r^2}[r-x.\frac{x}{r}]+\frac{x}{r}.f''(r).\frac{x}{r}\\=\frac{f'(r)}{r^3}(r^2-x^2)+\frac{f''(r)}{r^2}x^2 \\ \text{Similarly,}\,\, \frac{\partial^2u}{\partial y^2}=\frac{f'(r)}{r^3}(r^2-y^2)+\frac{f''(r)}{r^2}y^2\\ \text{and}\,\, \frac{\partial^2u}{\partial z^2}=\frac{f'(r)}{r^3}(r^2-z^2)+\frac{f''(r)}{r^2}z^2\\ \text{Now, adding we get,}\,\,\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2} \\=\frac{f'(r)}{r^3}[3r^2-(x^2+y^2+z^2)]+\frac{f''(r)}{r^2}(x^2+y^2+z^2)\\= \frac{f'(r)}{r^3}[3r^2-r^2]+\frac{ff''(r)}{r^2}.r^2\\=\frac{f'(r)}{r^3}. 2r^2+f''(r)\\=\frac{2}{r}f'(r)+f''(r)\\ \text{Second part.}\,\, f(r)=r^m \\ \therefore f'(r)=mr^{m-1} \\ \text{and}\,\, f''(r)=m(m-1)r^{m-2}. \\ \text{Now, L.H.S.}=\frac{2}{r}.f'(r)+f''(r)\\=\frac{2}{r}.mr^{m-1}+m(m-1)r^{m-2}\\=2mr^{m-2}+m(m-1)r^{m-2}\\=r^{m-2}[2m+m^2-m]\\=r^{m-2}(m+m^2)\\=m(m+1)r^{m-2}$

 4. If $\,\,z=f(x,y)\,\text{and}\,\, x=r \cos{\theta},\,\,y=r \sin{\theta}, \\ \text{Prove that,}\,\, \left( \frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=\left(\frac{\partial z}{\partial r}\right)^2\\+\frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2$

Sol. $\,\, \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}.\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}.\frac{\partial y}{\partial r}\\=\frac{\partial z}{\partial x}.\cos{\theta} +\frac{\partial z}{\partial y}. \sin{\theta} \cdots \cdots (1)\\ \frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}.\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}.\frac{\partial y}{\partial \theta}\\=\frac{\partial z}{\partial x}(-r \sin{\theta})+\frac{\partial z}{\partial y}.(r \sin{\theta})\\ \therefore \frac{1}{r}.\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}.(-\sin{\theta})+\frac{\partial z}{\partial y}.\cos{\theta} \cdots \cdots (2) \\$

Squaring (1) and (2), and adding them, 

 $\left(\frac{\partial z}{\partial r}\right)^2+\frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2\\=\left(\frac{\partial z}{\partial x} \cos{\theta} +\frac{\partial z}{\partial y} \sin{\theta} \right)^2\\+\left(-\frac{\partial z}{\partial x} \sin{\theta} +\frac{\partial z}{\partial y} \cos{\theta} \right)^2\\= \left( \frac{\partial z}{\partial x}\right)^2(\cos^2{\theta}+\sin^2{\theta})+\left( \frac{\partial z}{\partial y}\right)^2(\cos^2{\theta}\\+\sin^2{\theta})\\=\left( \frac{\partial z}{\partial x}\right)^2+\left( \frac{\partial z}{\partial y}\right)^2$

5. Solve (By Method of Seperation of Variables) $\,\,\frac{\partial u}{\partial x}= 2. \frac{\partial u}{\partial t}+u,\,\,\text{where,}\,\, u(x,0)=6e^{-3x}.$

Solution. Let the sol. of the given equation be : $u(x,t)= X(x) T(t). \\ \therefore x' T=2XT'+XT \\ \Rightarrow (X'-X)T=2XT' \\ \Rightarrow \frac{X'-X}{X}=2. \frac{T'}{T}=k(\neq 0), \text{say} \\ \therefore \frac{X'-X}{X}=k \\ \therefore \frac{X'}{X}=k+1 \\ \therefore \log{X}=(k+1)x+\log C_1 \\=(k+1)x \log{e} +\log{C_1}\,\,\,[\text{By Integrating both sides of the equation}]\\ \Rightarrow \log{X}=\log \{e^{x(k+1)}  \times C_1\}\\ \therefore X=C_1e^{x(k+1)} \\ \text{Solving,}\,\, \frac{T'}{T}=\frac{k}{2} \\ \therefore \log{T}=\frac{k}{2}. t+ \log{C_2}\\=\frac{kt}{2}\log{e}+\log{C_2} \\ \therefore \log{T}=\log{e^{kt/2}.C_2} \\ \therefore T=C_2.e^{kt/2} \\ \therefore u(x,t)=XT \\=C_1e^{(k+1)x}.C_2e^{kt/2}\\=(C_1C_2)e^{(k+1)x+kt/2} \\ \text{When,}\,\, t=0, u(x,0)=C_1C_2.e^{(k+1)x}=6e^{-3x}\,\,[\text{given}]$

On comparing, $\,\,C_1C_2=6,\,\,k+1=-3 \Rightarrow k=-4 \\ \therefore u(x,t)=6e^{-3x-4t/2}=6.e^{-3x-2t}=6e^{-(3x+2t)}$

6. If $\,\, x^n-1=(x^2-1) \prod_{k=1}^{(n-2)/2}  \left(x^2-2x\cos{\frac{2k\pi}{n}}+1 \right),$ if n be an even  positive integer.  Also, Deduce that $\,\,\sin{\frac{\pi}{32}}.\sin{\frac{2\pi}{32}}.\sin{\frac{3\pi}{32}}.\cdots \cdots \sin{\frac{15\pi}{32}}=\frac{1}{2^{13}}$

Sol. $\,\,\, x^n-1=0 \\ \Rightarrow x^n=1=\cos{2k\pi}+i\sin{2k \pi},\,\, k\,\text{ being an integer}\\ \Rightarrow x= \left(\cos{2k\pi}+i\sin{2k \pi} \right)^{1/n} =\cos{\frac{2k \pi}{n}}+i \sin{\frac{2k \pi}{n}}, \\ \text{where,}\,\, k=0,1,2,3, \cdots \cdots ,(n-1) \\ \therefore \,\, \text{ The n n-th roots of unity are  }\,\, \cos{\frac{2k \pi}{n}}+i \sin{\frac{2k \pi}{n}}; $

 They have the same modulus $\,1\,$ and their arguments are $\,\, 0,\frac{2\pi}{n},\frac{4\pi}{n}, ....,(2\pi- \frac{2 \pi}{n}). $   

Now,if $\,n\,$ be even, there, there are two real roots corresponding to $\,\,k=0,\frac n2,.$ 

The real roots are $\,\, 1,-1.\,$ Again , the roots corresponding to $\,k=r,\,\,$ and $\,k=n-r\,$    are : $\,\,\cos{\frac{2r \pi}{n}}+i \sin{\frac{2r\pi}{n}}\,\,$ respectively. 

So, when $\,n\,$ is even, the roots can be exhibited as $\quad \pm 1, \cos{\frac{2r \pi}{n}} \pm i \sin{\frac{2r\pi}{n}}. $                       

   Therefore   $x^n-1 \\= (x^2-1) \prod_{k=1}^{(n-2)/2}\left[  \left(x-\cos{\frac{2k \pi}{n}}-i \sin{\frac{2k\pi}{n}}\right) \\ \times \left(x-\cos{\frac{2k \pi}{n}}+i \sin{\frac{2k\pi}{n}}\right)\right] \\=(x^2-1) \prod_{k=1}^{(n-2)/2}  \left(x^2-2x\cos{\frac{2k\pi}{n}}+1 \right)$

2nd part: With $\,\,n=32,\,\,$ we can rewrite the equation like this.

$\frac{x^{32}-1}{x^2-1}=\prod_{k=1}^{15}(x^2+1-2x \cos{\frac{k\pi}{16}})\,\,\text{ for all x}\neq 0 \\ \therefore \lim_{x \to 1} \frac{x^{32}-1}{x^2-1}= \lim_{x \to 1} \prod_{k=1}^{15}(2-2\cos{\frac{k\pi}{16}})\cdots (1)$

So, L.H.S. of (1) is given by $\\ \lim_{x \to 1} \frac{x^{32}-1}{x^2-1}\\=\lim_{x \to 1} \frac{32x^{31}}{2x} \,\,[\text{By using L'Hospital's Law}]\\=\lim_{x \to 1} 16 x^{30}\\=16$ 

From (1), $16=\prod_{k=1}^{15}(2.2\sin^2{\frac{k\pi}{32}}) \\ \Rightarrow 4^2=4^{15}\prod_{k=1}^{15}\left(\sin{\frac{k\pi}{32}}\right)^2\\ \Rightarrow 4^{-13}= \prod_{k=1}^{15}\left(\sin{\frac{k\pi}{32}}\right)^2 \\ \Rightarrow \prod_{k=1}^{15}\left(\sin{\frac{k\pi}{32}}\right)=\frac{1}{2^{13}} $

 

7. Using $\,\,x^n-1=(x-1)\prod_{k=1}^{(n-1)/2} \{x^2-2x\cos{\frac{2k\pi}{n}}\\+1\}\,\,$   if n be odd positive integer, deduce that $\,\,\sin{\frac{\pi}{25}}.\sin{\frac{2\pi}{25}}.\sin{\frac{3\pi}{25}}.....\sin{\frac{12\pi}{25}}=\frac{5}{2^{12}}$

Solution. Put $\,\,n=25\,\,$. 

Then $\,\frac{x^{25}-1}{x-1}=\prod_{k=1}^{12} \{x^2-2x \cos{\frac{2k\pi}{25}}+1\} \\ \Rightarrow \lim_{x \to 1}\frac{x^{25}-1}{x-1}\\=\lim_{x \to 1} \prod_{k=1}^{12} \{x^2 -2x \cos{\frac{2k\pi}{25}}+1\}\\ \Rightarrow \lim_{x \to 1} \frac{25x^{24}}{1}=\prod_{k=1}^{12}\{2-2\cos{\frac{2k\pi}{25}}\} \\ \Rightarrow 25 \times 1=\prod_{k=1}^{12}2.2 \sin^2{\frac{k\pi}{25}} \\ \Rightarrow 5^2=\left(\prod_{k=1}^{12} 2 \sin{\frac{k\pi}{25}} \right)^2 \\ \therefore 5= \left(\prod_{k=1}^{12} 2 \sin{\frac{k\pi}{25}} \right) =2^{12}\prod_{k=1}^{12} \sin{\frac{k\pi}{25}} \\ \Rightarrow \frac{5}{2^{12}}=\prod_{k=1}^{12} \sin{\frac{k\pi}{25}}$

8. Show that the solutions of the equation $\,\,(1+x)^{2n+1}-(1-x)^{2n+1}=0\,\,$ are $\,\,x=0, \pm i \tan{\frac{r \pi}{2n+1}},r=1,2,3,....,n$

 Sol. $\,\, (1+x)^{2n+1}=(1-x)^{2n+1} \\ \Rightarrow \left( \frac{1+x}{1-x} \right)^{2n+1}=1=\cos{2r\pi}+i\sin{2r \pi}\\ \Rightarrow \frac{1+x}{1-x}=\cos{\frac{2r\pi}{2n+1}}+i \sin{\frac{2r\pi}{2n+1}},\,\, r=1,2,3,.....,n \\$

 $ \text{Hence,}\,\, \frac{2}{2x}= \frac{\cos{\frac{2r\pi}{2n+1}+i\sin{\frac{2r\pi}{2n+1}}}+1}{\cos{\frac{2r\pi}{2n+1}+i\sin{\frac{2r\pi}{2n+1}}}-1} \\ \quad [\text{By Componendo-dividendo}] \\ =\frac{2\cos^2{\frac{r\pi}{2n+1}+2i \sin{\frac{r\pi}{2n+1}}}\cos{\frac{r\pi}{2n+1}}}{-2\sin^2{\frac{r\pi}{2n+1}+2i \sin{\frac{r\pi}{2n+1}}}\cos{\frac{r\pi}{2n+1}}} \\= \frac{2 \cos{\theta}(\cos{\theta}+i\sin{\theta})}{2i \sin{\theta}(\cos{\theta}+i\sin{\theta})}\\ =\frac{1}{i}\cot{\theta} \\ \Rightarrow x= i\tan{\theta} $

LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS :

Procedure: Let it be required to find the stationary values of a function of 3 variables , say $\,\,u=f(x,y,z) \cdots (1)\\$ Subject to the condition, $\,\,\phi(x,y,z)=0 \cdots (2) \\ \text{and}\,\, \psi(x,y,z)=0 \cdots (3)$

For this we adopt the following procedure, construct the function :

$F=f+\lambda_1 \phi +\lambda_2 \psi \,\,\text{where}\,\, \lambda_1,\lambda_2$ are called non-zero multipliers.

From the equation $\frac{\partial F}{\partial x}=0,\frac{\partial F}{\partial y}=0,\frac{\partial F}{\partial z}=0, \cdots (4) $

Now, eliminating $\,\, x,y,z, \lambda_1,\lambda_2\,\,$ between the the equation (1) to (4), we are left with an equation in $\,u,\,$ the roots of which give the stationary values of $\,\,u \equiv f(x,y,z).$

9. Show that the stationary value of $\,\,a^3x^2+b^3y^2+c^3z^2\,\,\text{where}\, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\,$is given by $\,\,x=\frac{a+b+c}{a}, y=\frac{a+b+c}{b}, z=\frac{a+b+c}{c}$

Solution. Let $\, u \equiv f(x,y,z)=a^3x^2+b^3y^2+c^3z^2 \cdots (1)\\ \phi(x,y,z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1 =0 \cdots (2)$

Construct the function      

 $\,\,F=f+\lambda \phi ,\text{where}\,\, \lambda \,\,\text{is a non-zero multiplier} \\ F=(a^3x^2+b^3y^2+c^3z^2)+\lambda (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1) \\ \text{Now, from the equations, we get} \\  \frac{\partial F}{\partial x}=0,\,\, \frac{\partial F}{\partial y}=0,\,\,\frac{\partial F}{\partial z}=0 \\ \Rightarrow 2a^3x-\frac{\lambda}{x^2}=0 \Rightarrow 2a^3x^3=\lambda \cdots (3) \\ 2a^3y^3=\lambda \cdots (4) \\ 2a^3z^3=\lambda \cdots (5)\\ \text{From (3),(4) and (5) we get,}\,\,\,\, 2a^3x^3=2b^3y^3\\=2c^3z^3 \\ \Rightarrow ax=by=cz =t \,\,[\text{say}] \\ \therefore x=\frac{t}{a}; y=\frac{t}{b}; z=\frac{t}{c}$ 

Substituting these in (2), we get  $\,\, \frac{a+b+c}{t}=1 \Rightarrow t=a+b+c; \\ \text{From (6),}x=\frac{a+b+c}{a}; y=\frac{a+b+c}{b}; z=\frac{a+b+c}{c} $ 

So, these are the required values of $x,y,z$ for $u$ to have stationary values.

10. Find the stationary values of  $x^2+y^2+z^2\,\,$ subject to the conditions $\,\,ax^2+by^2+cz^2+2fyz+2gzx+2hxy=1  \\ \text{and}\,\, lx+my+nz=0$

Sol. Let $\,\,u=x^2+y^2+z^2  \cdots (1)\\ \phi = ax^2+by^2+cz^2+2fyz+2gzx\\+2hxy-1=0 \cdots (2)\\ \psi= lx+my+nz=0 \cdots (3)\\ \text{Construct the function} \,\,F= u+\lambda_1\phi +\lambda_2\psi \\ \Rightarrow F=(x^2+y^2+z^2) \\ +\lambda_1(ax^2+by^2+cz^2+2fyz \\+2gzx+2hxy-1)+\lambda_2(lx+my+nz) \\ \text{From the equations, we get}\,\, \frac{\partial F}{\partial x}=0,\,\, \frac{\partial F}{\partial y}=0,\,\,\frac{\partial F}{\partial z}=0 \\ \Rightarrow 2x+2\lambda_1(ax+gz+hy)+\lambda_2l=0 \cdots (4)\\ 2y+2\lambda_1(by+hx+fz)+\lambda_2m=0 \cdots (5)\\ 2z+2\lambda_1(cz+fy+gx)+\lambda_2n=0 \cdots (6)$

Taking $(4) \times x + (5) \times y+ (6)\times z, \text{we get} \\ 2(x^2+y^2+z^2) +2\lambda_1(ax^2+by^2+cz^2+2fyz \\+2gzx+2hxy)+\lambda_2(lx+my+nz)=0 \\ \Rightarrow 2u+2\lambda_1(1)+\lambda_2(0)=0 \\ \Rightarrow\lambda_2=-u \\ \text{Substituting in (4),(5),(6) we get,}\\ (1-ax)x-huy-guz+\lambda_2l=0  \cdots (7A) \\ -hux+(1-bu)y-fuz+\lambda_2m=0 \cdots (7B)\\ -gux-fuy+(1-cu)z+\lambda_2n=0 \cdots (7C) \\ \text{Also from (3),} \,\,lx+my+nz=0 \cdots (7D) $

Eliminating x,y,z and $\,\,\lambda_2\,\,$ from equation (7) by Cramer's rule,

 $\begin{vmatrix} 1-au&-hu &-gu &l \\ -hu& 1-bu &-fu &m \\ -gu& -fu & 1-cu &n \\ l& m & n & 0 \end{vmatrix}=0$, the roots of which gives the stationary values of $u.$

In the next article, we will discuss few more mathematical problems and their solutions , which may be beneficial for students who are preparing for wbcs math optional papers.