Such a box has one vertex at the origin (0, 0, 0), and the vertex opposite this one is affixed to the plane with coordinates [tex](x,y,z)[/tex]. The volume of the box is then [tex]xyz[/tex], which we want to maximize subject to the constraint [tex]x+2y+3z=15[/tex]. Of course, we don't want a degenerate box, so we assume each of [tex]x,y,z[/tex] is positive.
We can use Lagrange multipliers - the Lagrangian is
[tex]L(x,y,z)=xyz+\lambda(x+2y+3z-15)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=yz+\lambda=0[/tex]
[tex]L_y=xz+2\lambda=0[/tex]
[tex]L_z=xy+3\lambda=0[/tex]
[tex]L_\lambda=x+2y+3z-15=0[/tex]
Then
[tex]yz+\lambda=0\implies\lambda=-yz[/tex]
[tex]xz+2\lambda=0\implies xz-2yz=z(x-2y)=0\implies x=2y[/tex]
[tex]xy+3\lambda=0\implies xy-3yz=y(x-3z)=0\implies x=3z[/tex]
[tex]x+2y+3z-15=0\implies x+x+x=15\implies x=5\implies y=\dfrac52\text{ and }z=\dfrac53[/tex]
So the largest volume that can be attained is [tex]5\cdot\dfrac52\cdot\dfrac53=\dfrac{125}6[/tex].
The volume of a box is the amount of space in it.
The volume of the largest rectangular box is 125/6
The vertex of the plane is given as:
[tex]\mathbf{x + 2y + 3z = 15}[/tex]
Subtract 15 from both sides
[tex]\mathbf{x + 2y + 3z -15= 0}[/tex]
Using Lagrange multipliers, we have:
[tex]\mathbf{L(x,y,z) = xyz + \lambda(0)}[/tex]
Substitute [tex]\mathbf{x + 2y + 3z -15= 0}[/tex]
[tex]\mathbf{L(x,y,z) = xyz + \lambda(x + 2y + 3z- 15)}[/tex]
Differentiate
[tex]\mathbf{L_x = yz + \lambda \times 1}[/tex]
[tex]\mathbf{L_x = yz + \lambda}[/tex]
[tex]\mathbf{L_y = xz + \lambda \times 2}[/tex]
[tex]\mathbf{L_y = xz + 2\lambda }[/tex]
[tex]\mathbf{L_x = xy + \lambda \times 3}[/tex]
[tex]\mathbf{L_y = xy + 3\lambda }[/tex]
[tex]\mathbf{L_{\lambda} = 0 + x + 2y + 3z - 15}[/tex]
[tex]\mathbf{L_{\lambda} = x + 2y + 3z - 15}[/tex]
Equate the above derivatives o 0
[tex]\mathbf{yz + \lambda = 0}[/tex]
[tex]\mathbf{xz + 2\lambda = 0}[/tex]
[tex]\mathbf{xy + 3\lambda = 0}[/tex]
[tex]\mathbf{x + 2y + 3z -15= 0}[/tex]
Make [tex]\mathbf{\lambda}[/tex] the subject in [tex]\mathbf{yz + \lambda = 0}[/tex]
[tex]\mathbf{\lambda = -yz}[/tex]
Substitute [tex]\mathbf{\lambda = -yz}[/tex] in [tex]\mathbf{xz + 2\lambda = 0}[/tex] and [tex]\mathbf{xy + 3\lambda = 0}[/tex]
[tex]\mathbf{xz - 2yz = 0}[/tex]
[tex]\mathbf{xy - 3yz = 0}[/tex]
Factor out z and y
[tex]\mathbf{z (x - 2y) = 0}[/tex]
[tex]\mathbf{y (x - 3z) = 0}[/tex]
Divide both sides by z
[tex]\mathbf{x - 2y = 0}[/tex]
[tex]\mathbf{x - 3z = 0}[/tex]
Make x the subject
[tex]\mathbf{x = 2y}[/tex]
[tex]\mathbf{x = 3z}[/tex]
Substitute [tex]\mathbf{x = 2y}[/tex] and [tex]\mathbf{x = 3z}[/tex] in [tex]\mathbf{x + 2y + 3z = 15}[/tex]
[tex]\mathbf{x + x + x =15}[/tex]
[tex]\mathbf{3x =15}[/tex]
Divide by 3
[tex]\mathbf{x =5}[/tex]
Substitute [tex]\mathbf{x =5}[/tex] in [tex]\mathbf{x = 2y}[/tex] and [tex]\mathbf{x = 3z}[/tex]
[tex]\mathbf{5 = 2y}[/tex]
[tex]\mathbf{y = \frac52}[/tex]
[tex]\mathbf{5 = 3z}[/tex]
[tex]\mathbf{z = \frac 53}[/tex]
The volume of the largest rectangular box is then calculated as:
[tex]\mathbf{V_{max} = xyz}[/tex]
So, we have:
[tex]\mathbf{V_{max} = 5 \times \frac 52 \times \frac53}[/tex]
[tex]\mathbf{V_{max} = \frac{125}{6}}[/tex]
Hence, the volume of the largest rectangular box is 125/6
Read more about volumes at:
https://brainly.com/question/13499875
