Maximize volume of box using 2nd derivative test

Question

The equation of the plane is $ x/6+y/6+z/6=1$ .

I'm lost on how to find the dimensions of the box with maximum volume. I know we use the second derivative test and make sure the second derivative test is greater than 0 and check boundary points.

Answer 1

Alternative method :using AM.GM inequality $$\frac{a+b+c}{3}\geq\sqrt[3]{abc}$$ you have $$\frac{x+y+z}{6}=1 \to x+y+z=6 \\\sqrt[3]{xyz}\leq \frac{x+y+z}{3}$$ note that :volume of cube is $x\times y \times z$ so $$V_{max} \leq (\frac{x+y+z}{3})^3\\ V_{max} \leq (\frac{6}{3})^3=8\\\to V_{max}=8$$ Hint for your method :
$$v=x.y.z \to v=\underbrace{x.y.z}_{(\frac{x+y+z}{6}=1)}=xy(6-x-y)\\v=6xy-x^2y-xy^2\\v=f(x,y)=6xy-x^2y-xy^2 \to f_{xx}f_{yy}-f_{xy}^2|_{(0,0)}\\f_{xx}=2y,f_{yy}=-2x ,fxy=6-2x-2y\\f_{xx}f_{yy}-f_{xy}^2|_{(0,0)}=0.0-6^2$$