Find the dimensions for the maximum volume of a cuboid under this slightly strange condition.
It has its total length of edges equal to the square of the space diagonal length.
I am having trouble with this question as I can't seem to get anywhere with the x, y and z dimensions. So far I have said that:
x is the width of the box
y is the depth of the box
z is the height of the box
the square of the space diagonal $d^2 = x^2+y^2+z^2$
the perimeter would be $4x+4y+4z$
Volume of cuboid is $xyz$
Using the lagrange multiplier formula $$L=xyz-\lambda(x^2+y^2+z^2-4x-4y-4z)$$ $$L_x = yz-\lambda(2x-4) =0 \Rightarrow \lambda = \frac {yz}{2x-4}$$ $$L_y = xz-\lambda(2y-4) =0 \Rightarrow \lambda = \frac {xz}{2y-4}$$ $$L_y = xy-\lambda(2z-4) =0 \Rightarrow \lambda = \frac {xy}{2z-4}$$
The problem I'm having is when I make the $\lambda$ equal to each other, I can't seem to find the dimensions. Any advice or hints would be appreciated.
Thanks.
no the mistake you have done is using $\lambda$ alone.Use two parameters.Say $xyz-\lambda_1(x^2+y^2+z^2)-\lambda_2(4x+4y+4z)=0$.