Question

If diagonal of a box is fixed = 1 unit; and length = x , breadth = y, height = z. Then; using Language Multiplier method, the problem : maximise the volume of the box":

Answer 1

$\text{We need to maximize: v(x,y,z)=xyz}\\ \text{Subject to: g(x,y,z)}=x^2+y^2+z^2-1=0\\ \text{By equating }\nabla \text{v(x,y,z)}=\text{yzi}+\text{xzj}+\text{xyk} \text{ and } \lambda \nabla\text{g(x,y,z)}=\lambda\text{(2xi}+\text{2yj}+\text{2zk) we obtain the following system of equations:}\\ yz=2x\lambda\\ xz=2y\lambda\\ xy=2z\lambda\text{ and}\\ x^2+y^2+z^2=1\\ \text{Solving this system yields the following real solutions: }\\ (x,y,z)=(-1,0,0)\\ (x,y,z)=(0,-1,0)\\ (x,y,z)=(0,0,-1)\\ (x,y,z)=(0,0,1)\\ (x,y,z)=(0,1,0)\\ (x,y,z)=(1,0,0)\\ (x,y,z)=(-\frac{\sqrt3}{3},-\frac{\sqrt3}{3},-\frac{\sqrt3}{3})\\ (x,y,z)=(-\frac{\sqrt3}{3},-\frac{\sqrt3}{3},\frac{\sqrt3}{3})\\ (x,y,z)=(-\frac{\sqrt3}{3},\frac{\sqrt3}{3},-\frac{\sqrt3}{3})\\ (x,y,z)=(-\frac{\sqrt3}{3},\frac{\sqrt3}{3},\frac{\sqrt3}{3})\\ (x,y,z)=(\frac{\sqrt3}{3},-\frac{\sqrt3}{3},-\frac{\sqrt3}{3})\\ (x,y,z)=(\frac{\sqrt3}{3},-\frac{\sqrt3}{3},\frac{\sqrt3}{3})\\ (x,y,z)=(\frac{\sqrt3}{3},\frac{\sqrt3}{3},-\frac{\sqrt3}{3})\\ (x,y,z)=(\frac{\sqrt3}{3},\frac{\sqrt3}{3},\frac{\sqrt3}{3})\\ \text{Now, substituting the above solutions into v(x,y,z) yields:}\\ \text{v}(-1,0,0)=0\\ \text{v}(0,-1,0)=0\\ \text{v}(0,0,-1)=0\\ \text{v}(0,0,1)=0\\ \text{v}(0,1,0)=0\\ \text{v}(1,0,0)=0\\ \text{v}(-\frac{\sqrt3}{3},-\frac{\sqrt3}{3},-\frac{\sqrt3}{3})=-\frac{\sqrt3}{9}\\ \text{v}(-\frac{\sqrt3}{3},-\frac{\sqrt3}{3},\frac{\sqrt3}{3})=\frac{\sqrt3}{9}\\ \text{v}(-\frac{\sqrt3}{3},\frac{\sqrt3}{3},-\frac{\sqrt3}{3})=\frac{\sqrt3}{9}\\ \text{v}(-\frac{\sqrt3}{3},\frac{\sqrt3}{3},\frac{\sqrt3}{3})=-\frac{\sqrt3}{9}\\ \text{v}(\frac{\sqrt3}{3},-\frac{\sqrt3}{3},-\frac{\sqrt3}{3})=\frac{\sqrt3}{9}\\ \text{v}(\frac{\sqrt3}{3},-\frac{\sqrt3}{3},\frac{\sqrt3}{3})=-\frac{\sqrt3}{9}\\ \text{v}(\frac{\sqrt3}{3},\frac{\sqrt3}{3},-\frac{\sqrt3}{3})=-\frac{\sqrt3}{9}\\ \text{v}(\frac{\sqrt3}{3},\frac{\sqrt3}{3},\frac{\sqrt3}{3})=\frac{\sqrt3}{9}\\ \text{Thus the maximum volume of the box is }\frac{\sqrt3}{9}.$