the box's volume is #V = xyz# but we know that #z = 4-x^2-y^2#. i'll stuff a 2-D plot of that in at the end, it's symmetrical as it's rotated about the z axis
we can do this 2 ways, first just by re-writing the volume in 2 variables as
#V = xyz = xy(4-x^2-y^2)#
#= 4xy-x^3 y-xy^3#
we can then look for critical points using the partial derivatives
#V_x= 4y-3x^2 y-y^3 = y (4-3x^2-y^2)#
and
#V_y= 4x-x^3 -3xy^2 = x(4-x^2 -3y^2)#
The trivial solution is at (0,0) when the volume is zero. We can ignore that for physical reasons.
but #V_x = 0 implies 4-3x^2-y^2 = 0#
and #V_y= 0 implies 4-x^2 -3y^2 = 0#
combining these we have
#4-3(4 - 3y^2)-y^2 = 0#
#implies 8y^2= 8# which in the first quad means #y = 1# and by symmetry #x = 1# which gives us #z = 2# so that #V = 1*1*2 = 2#
in order to explore whether this is actually a max, we can look to the Hessian matrix: #((V_(x x), V_(xy)), (V_(yx), V_(yy)))_(1,1) #. But this looks like an awful lot of algebra, or we can again look at physical arguments. I might edit this answer later to include some more stuff on this aspect.
but for now we can do this a second way, by using the method of Lagrange Multipliers.
We have function
#V = xyz# to optimise in light of constraint #C(x,y,z) = x^2 + y^2 + z = 4 = const#
so #nabla V = lambda nabla C#, with #lambda# as the multiplier, gives us
#((yz),(xz),(xy)) = lambda ((2x),(2y),(1))#
or #lambda = color(red)((yz)/(2x)) = color(green)((xz)/(2y)) = color(blue)((xy)/(1))#
from green = blue, we have #xz = 2xy^2# and leaving aside the trivial #x=0# solution we have #z = 2y^2#
from red = blue we have: #yz = 2x^2 y#, and leaving aside the trivial #x=0# solution we have #z = 2x^2#
we combine those as #x^2 = y^2#, and as we are in the first octant we can say that #y = x#
The constraint equation becomes #C(x,y,z) = x^2 + x^2 + 2x^2 = 4 implies 4x^2 = 4# and in the first octant #x = 1# so #y=1#, #z = 2# and #V = 2#.
Same as before.
graph{4-x^2 [-10, 10, -5, 5]}
