Respuesta :
Considering the spaces in the coefficients, I'm guessing the plane has equation
[tex]x+\dfrac13y+\dfrac18z=1[/tex]
Let [tex](x,y,z)[/tex] be the vertex of the box that lies in this plane. The box has volume
[tex]V(x,y,z)=xyz[/tex]
and we want to maximize this subject to the plane equation. Using the method of Lagrange multipliers, the Lagrangian is
[tex]L(x,y,z,\lambda)=xyz+\lambda\left(x+\dfrac13y+\dfrac18z-1\right)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=yz+\lambda=0[/tex]
[tex]L_y=xz+\dfrac13\lambda=0[/tex]
[tex]L_z=xy+\dfrac18\lambda=0[/tex]
[tex]L_\lambda=x+\dfrac13y+\dfrac18z-1=0[/tex]
We have
[tex]xL_x+yL_y+zL_z=3xyz+\lambda\left(x+\dfrac13y+\dfrac18z\right)=0[/tex]
[tex]L_\lambda=0\implies3xyz+\lambda=0[/tex]
[tex]L_x=0\implies3xyz-yz=yz(3x-1)=0\implies x=\dfrac13[/tex]
(we ignore [tex]y=0[/tex] and [tex]z=0[/tex] because those values correspond to a degenerate box with zero volume)
[tex]L_y=0\implies3xyz-3xz=3xz(y-1)=0\implies y=1[/tex]
[tex]L_z=0\implies3xyz-8xy=xy(3z-8)=0\implies z=\dfrac83[/tex]
So we get a maximum volume of [tex]V\left(\dfrac13,1,\dfrac83\right)=\dfrac89[/tex].
Using the contour pictures can help getting the limits for integration which then gives the needed volume. The maximum volume inside the given box is 1/1404 cubic units approx.
How to find the volume of a region?
Try getting its infinitesimal volume, then integrate it over the region of interest. That will provide you with the volume of the considered region.
The considered region is enclosed by the planes:
Coordinate planes X-Y, Y-Z, and Z-X planes.
The plane [tex]x + 13y + 18z = 1[/tex]
The area bounded by them lies in the all axes positive octant.
Intersections on x axis is when y and z are 0, thus:
[tex]x + 13y + 18z = 1\\x + 0 + 0 = 1\\x = 1[/tex]
Thus, intersection on x-axis is on the point (1,0,0)
Similarly, we get: intersection on y-axis at point (0, 1/13, 0),
and intersection on z-axis at point (0, 0 , 1/18)
Now, if we cut the 3D model parallel to X-Y plane at particular height, say Z = z, we get the equation of line on the plane we got from slice as:
[tex]x + 13y = 1-18z[/tex] (take right side as a constant, as we're on a specific z).
The area of the triangle made between this line and X and Y axis line on that same sliced plane is obtained by same slicing method.
Taking a point on y axis, then its length from y axis to curve is [tex]x = 1-18z - 13y[/tex]
Multiplying this length with dy will give infinitesimal rectangle's area, and integrating it over y axis from 0 to (1-18z)/13 (putted x =0 to get its upper limit), will give us the needed area:
Area of triangle = [tex]\int_0^{(1-18z)/13}(1-18z - 13y)dy[/tex]
This area times very small unit from z dimension (dz) gives volume of infinitesimal slice of that interested region as:
Volume of infinitesimally thin slice =[tex]Area \times dz =(\int_0^{(1-18z)/13}(1-18z - 13y)dy )dz[/tex]
Integrating it over z axis(from 0 to 1/18, we get the needed volume as:
[tex]V =\int_0^{1/18}(\int_0^{(1-18z)/13}(1-18z - 13y)dy )dz\\V = \int_0^{1/18}[y-18zy-13y^2/2]_0^{(1-18z)/13}dz\\\\V = \int_0^{1/18}(1-18z)^2(\dfrac{1}{26})dz\\\\V =\dfrac{1}{26} [z + 364z^3/3 - 36z^2/2]_0^{1/18}\\V = \dfrac{1}{26} (\dfrac{1}{18} + \dfrac{1}{54} - \dfrac{1}{18}) = \dfrac{1}{1404}[/tex]
Thus, The maximum volume inside the given box is 1/1404 cubic units approx.
Learn more about volume here:
https://brainly.com/question/6957600
