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Kruka [31]
3 years ago
10

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

he plane x+2y+3z=6.
Mathematics
1 answer:
nataly862011 [7]3 years ago
4 0

Let (<em>x</em>, <em>y</em>, <em>z</em>) be a point on the plane in the first octant. The box formed by this point has volume <em>xyz</em>, and you want to maximize this subject to the equation of the plane.

Use the method of Lagrange multipliers: the Lagrangian is

<em>L</em>(<em>x</em>, <em>y</em>, <em>z</em>) = <em>xyz</em> - <em>λ</em> (<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6)

Find its critical points:

∂<em>L</em>/∂<em>x</em> = <em>yz</em> - <em>λ</em> = 0

∂<em>L</em>/∂<em>y</em> = <em>xz</em> - 2<em>λ</em> = 0

∂<em>L</em>/∂<em>z</em> = <em>xy</em> - 3<em>λ</em> = 0

∂<em>L</em>/∂<em>λ</em> = -(<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6) = 0

Solving the first three equations for <em>λ</em> gives

<em>λ</em> = <em>yz</em> = <em>xz</em>/2 = <em>xy</em>/3

Solve these equations for <em>y</em> and <em>z</em> :

• <em>yz</em> = <em>xz</em>/2   =>   <em>y</em> = <em>x</em>/2   =>   2<em>y</em> = <em>x</em>

• <em>yz</em> = <em>xy</em>/3   =>   <em>z</em> = <em>x</em>/3   =>   3<em>z</em> = <em>x</em>

Substitute these solutions into the last equation and solve for <em>x</em>, then again for <em>y</em> and <em>z</em> :

<em>x</em> + 2<em>y</em> + 3<em>z</em> - 6 = 3<em>x</em> - 6 = 0   =>   3<em>x</em> = 6   =>   <em>x</em> = 2, <em>y</em> = 1, <em>z</em> = 2/3

At this critical point, the maximum volume is

<em>xyz</em> = 2*1*2/3 = 4/3

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