## Calculus: Early Transcendentals 8th Edition

$\dfrac{L^3}{3\sqrt 3}$
Use Lagrange Multipliers Method: $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ Volume of a box is given by $f(x,y,z)=V=xyz$ This yields $\nabla f=\lt yz,xz,xy \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z \gt$ Using the constraint condition we get, $yz=\lambda 2x, xz=\lambda 2y,xy=\lambda 2z$ After solving, we get $x=y=z$ Since, $g(x,y,z)=x^2+y^2+z^2=L^2$ yields $x^2+x^2+x^2=L^2$ $x=y=z=\dfrac{L}{\sqrt 3}$ Hence, Volume of a box is given by $f(x,y,z)=V=xyz=(\dfrac{L}{\sqrt 3})^3=\dfrac{L^3}{3\sqrt 3}$