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Find an equation of the plane that passes through the point (1, 3, 7) and cuts off the smallest volume in the first octant.

1 Answer

6 votes

Answer:

21x +7y +3z = 63

Explanation:

The equation of a plane in intercept form is ...

x/a +y/b +z/c = 1

where a, b, c are the x-, y-, and z-intercepts, respectively. The volume in the first octant cut off by this plane will be given by ...

V = (1/6)abc

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setup

The problem is to minimize the volume subject to the constraint that the plane go through the point. That is, our minimization problem is ...

minimize abc/6

subject to 1/a +3/b +7/c = 1

solution

As with many optimization problems, the optimum solution splits the constraint into equal parts: 1/3 +1/3 +1/3 = 1. That is ...

1/a = 3/b = 7/c = 1/3 ⇒ a = 3, b = 9, c = 21

The intercept form equation for the plane is ...

x/3 +y/9 +z/21 = 1

In standard form, this is ...

21x +7y +3z = 63

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Additional comment

The optimization problem can be solved using the method of Lagrange multipliers.

Th Lagrangian is ...

ℒ = (1/6)abc +λ(bc +3ac +7ab -abc)

and the partial derivatives are ...

∂ℒ/∂a = bc/6 +λ(3c +7b -bc) = 0

∂ℒ/∂b = ac/6 +λ(c +7a -ac) = 0

∂ℒ/∂c = ab/6 +λ(b +3a -ab) = 0

∂ℒ/∂λ = bc +3ac +7ab -abc = 0

Solving these simultaneous equations gives (a, b, c, λ) = (3, 9, 21, 1/2).

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