A plane with equation xa+yb+zc=1 (a, b,c>0)together with the positive coordinate planes forms a tetrahedron of volume V=16abcFind the plane that minimizes V if the plane is constrained to pass through a point P=(2,1,1) .

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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).

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

the system of inequalities is

step-by-step explanation:

step 1

find the equation of the dashed line

we have the points

(0,0) and (3,1)

the line passes through the origin, so, is a direct variation

find the slope

the equation of the dashed line is

the solution of the inequality a s the shaded area above the dashed line

therefore

the equation of the inequality a is

step 2

find the equation of the solid line

we have the points

(4,0) and (0,4)

find the slope

the equation of the solid line is

the solution of the inequality b s the shaded area below the solid line

therefore

the equation of the inequality b is

step 3

the system of inequalities is

i believe the answer is c.