**By using the method of Lagrange, we will get:**

**\(\displaystyle y=\lambda x\) ... (1)**

**\(\displaystyle 4x-3y=\lambda y\) ... (2)**

**\(\displaystyle 4=\lambda z\) ... (3)**

**\(\displaystyle 4x^2+y^2+4z^2=24\) ... (4)**

**By substituting (1) in (2), we will get the equation : \(\displaystyle x(-\lambda^2 - 3\lambda + 4)=0\)**

**\(\displaystyle \implies x=0,\lambda=-4,\lambda=1\) ..**

**when x=0 ---> y=0 , substitute this in 4 ---> \(\displaystyle z^2=6 \implies z=\pm \sqrt{6}\)**

**so the first two points are \(\displaystyle (0,0,\sqrt{6})\) and \(\displaystyle (0,0,-\sqrt{6})\)**

**when \(\displaystyle \lambda = 1\) ---> z=4 from (3) ---> y=x from (1) and (2)**

**So when \(\displaystyle \lambda=1\) we have y=x and z=4 , By substituting t****his in (4) :**

**\(\displaystyle 5x^2+64=24 \implies x^2=-8\) which don't have any solutions**

**so when \(\displaystyle \lambda = 1\), we do not have any points.**

**when \(\displaystyle \lambda=-4\) ---> y=-4x from (1) and z=-1 from (3)**

**by substituting this in (4) ---> \(\displaystyle x^2=1 \implies x=\pm 1\)**

**when x=1 --> y=-4 ---> third point is (1,-4,-1)**

**when x=-1 ---> y=4 ---> fourth point is (-1,4,-1)**

**The points are : \(\displaystyle (0,0,\sqrt{6}),(0,0,-\sqrt{6}),(1,-4,-1)\) and \(\displaystyle (-1,4,-1)\)**

**Find the value of the function at these point, biggest --> max, smallest --> min.**