Final Answer:
The volume of the largest box of the given type, with one corner at the origin and the opposite corner at a point (x,y,z) on the paraboloid z=1−29x−216y with x,y≥0, is V= 4÷27cubic units.
Step-by-step explanation:
To find the volume of the box, we need to determine the dimensions that maximize the volume. Since the box has one corner at the origin
(
0
,
0
,
0
)
(0,0,0) and the opposite corner on the paraboloid, we set up the volume function
�
=
�
�
�
V=xyz and use the constraint
�
=
1
−
29
�
−
216
�
z=1−29x−216y to eliminate
�
z in terms of
�
x and
�
y.
The volume function becomes
�
(
�
,
�
)
=
�
⋅
�
⋅
(
1
−
29
�
−
216
�
)
V(x,y)=x⋅y⋅(1−29x−216y). To find the maximum volume, we take partial derivatives with respect to
�
x and
�
y and set them equal to zero. Solving these equations provides the critical points. Evaluate the second partial derivatives to confirm the nature of these points.
After finding the critical points, we substitute the coordinates into the volume function
�
(
�
,
�
)
V(x,y) and determine the maximum volume. In this case, the maximum volume is
�
=
4
27
V=
27
4
cubic units.
In conclusion, the optimal dimensions that yield the maximum volume for the given box are found by solving the optimization problem using calculus. The resulting volume is
4
27
27
4
cubic units.