A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant of the paraboloid #z = 4-x^2-y^2#, what is the box's maximum volume?

Question

Cameron Alexander · Accepted Answer

2 cubic units

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]}

Paisley Johnson · Answer

undefined To find the maximum volume of the rectangular box, we need to maximize the product of its three dimensions: length, width, and height.

Let's denote the dimensions of the box as $ x $, $ y $, and $ z $. Since the box has three of its faces on the coordinate planes, its vertices are at $ (0,0,0) $, $ (x,0,0) $, $ (0,y,0) $, and $ (0,0,z) $.

The volume of the box is given by $ V = xyz $.

Given that one vertex of the box lies in the first octant of the paraboloid $ z = 4 - x^2 - y^2 $, we have the constraint that $ z = 4 - x^2 - y^2 $.

We want to maximize $ V = xyz $ subject to the constraint $ z = 4 - x^2 - y^2 $.

Using the constraint, we can express $ z $ in terms of $ x $ and $ y $ as $ z = 4 - x^2 - y^2 $.

Now, we have $ V = xy(4 - x^2 - y^2) $.

To maximize $ V $, we take the partial derivatives of $ V $ with respect to $ x $ and $ y $, set them equal to zero, and solve for $ x $ and $ y $.

Taking partial derivatives:

$ \frac{\partial V}{\partial x} = y(4 - 3x^2 - y^2) $

$ \frac{\partial V}{\partial y} = x(4 - x^2 - 3y^2) $

Setting them equal to zero:

$ y(4 - 3x^2 - y^2) = 0 $

$ x(4 - x^2 - 3y^2) = 0 $

The solutions to these equations occur when either $ x = 0 $, $ y = 0 $, or both terms in parentheses equal zero.

Solving these equations, we find the critical points to be $ (0, 0) $, $ (\pm \sqrt{3}, 0) $, and $ (0, \pm \sqrt{3}) $.

To find which of these points gives the maximum volume, we can use the second partial derivative test or evaluate the volume at each critical point.

However, we notice that the function $ V = xy(4 - x^2 - y^2) $ is symmetric about the $ x $-axis and $ y $-axis, so we can restrict our search to the first quadrant.

Evaluating the volume at $ (0, 0) $, $ (\sqrt{3}, 0) $, and $ (0, \sqrt{3}) $, we find that the maximum volume occurs at $ (\sqrt{3}, \sqrt{3}) $ and its symmetric points.

Therefore, the maximum volume of the rectangular box is achieved when $ x = y = \sqrt{3} $, and $ z = 4 - (\sqrt{3})^2 - (\sqrt{3})^2 = 4 - 3 = 1 $.

Thus, the maximum volume of the box is $ V = \sqrt{3} \times \sqrt{3} \times 1 = 3 $.