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A plane with equation (x/a) + (y/b) + (z/c) = 1, where a,b,c > 0 together with the positive coordinate planes form a tetrahedron of volume V = (1/6)abc. Find the plane that minimizes V if the plane is constrained to pass through the point P(2,1,1).
Answer:
The plane is x/6 + y/3 + z/3 = 1
Step-by-step explanation:
Given
Equation: (x/a) + (y/b) + (z/c) = 1 where a,b,c > 0
Minimise, V = (1/6) abc subject to
the constraint g = 2/a + 1/b + 1/c = 1
First, we need to expand V
V = (abc)/6
Possible combinations of V taking 2 constraints at a time; we have
(ab)/6, (ac)/6 and (bc)/6
Applying Lagrange Multipliers on the possible combinations of V, we have:
∇V = λ∇g
This gives
<bc/6, ac/6, ab/6> = λ<-2/a², -1/b², -1/c²>
If we equate components on both sides, we get:
(a²)bc/12 = -λ = a(b²)c/6 = ab(c²)/6
Solving for a, b and c;
First, let's equate:
(a²)bc/12 = a(b²)c/6 -- divide through by abc, we have
a/12 = b/6 --- multiply through by 12
12 * a/12 = 12 * b/6
a = 2 * b
a = 2b
Then, let's equate:
(a²)bc/12 = ab(c²)/6 -- divide through by abc, we have
a/12 = c/6 --- multiply through by 12
12 * a/12 = 12 * c/6
a = 2 * c
a = 2c
Lastly, we equate:
a(b²)c/6 = ab(c²)/6 -- divide through by abc, we have
b/6 = c/6 --- multiply through by 6
6 * b/6 = 6 * c/6
b = 2
Writing these three results, we have
a = 2b; a = 2c and b = c
Recalling the constraints;
g = 2/a + 1/b + 1/c = 1
By substituton, as have
2/(2c) + 1/c + 1/c = 1
1/c + 1/c + 1/c = 1
3/c = 1
c * 1 = 3
c = 3
Since a = 2c;
So, a = 2 * 3
a = 6
Similarly, b = c
So, b = 3
So, the plane: (x/a)+(y/b)+(z/c)=1;
By substituton, we have
x/6 + y/3 + z/3 = 1
Hence, the plane
So the plane is x/6 + y/3 + z/3 = 1
