Explain your answers in the following.

Question

(a) Show that the minimum of x+y+z subject to xyz = 1,x ≥ 0,y ≥ 0,z ≥ 0is3. Be sure to show why the minimum exists. Suggestion: Use the gradients to find the critical point, on the constraint surface, and the minimum point must exist and be in the interior of the cube [0, 3] × [0, 3] × [0, 3].

(b) Show that the minimum of (xy)^2 + (yz)^2 + (zx)^2 subject to xyz=1,x≥0,y≥0,y≥0 is 3. Suggestion: Use Part (a).

(c) Find the minimum area of the triangle determined by (a,0,0),(0,b,0),(0,0,c),subject to a ≥ 0,b ≥ 0,c ≥ 0,abc = 1. Hint: Use the cross product and Part (b).

Answer 1

(a)

Let

f(x, y, z) = x + y + z

g(x y, z) = xyz - 1

(x0, y0, z0) be a minimum of f(x, y, z) subject to g(x, y , z) = 0 (to be found)

We use the Lagrange multiplier λ and form the function

F(x, y, z, λ) = f(x, y, z) - λg(x, y, z) = x + y + z - λxyz - λ

To find critical points we get four equations with four unknowns

0 = xyz - 1

0 = ∂F/∂x = 1 - λyz = 1 - λ/x

0 = ∂F/∂y = 1 - λxz = 1 - λ/y

0 = ∂F/∂x = 1 - λxy = 1 - λ/z

The solution is x = y = z = λ = 1

So we have a critical point (1, 1, 1).

f(1, 1, 1) = 3.

So any minimum must satisfy

f(x0, y0, z0) <= 3. (1)

Next we show that no minimum (x0, y0, z0) can lie outside [0,3]×[0,3]×[0,3].

Assume to the contrary that, say, x0 > 3.

In order to satisfy (1), at least one of y0, z0 would have to be negative,

which would place them outside of the allowed range.

Next we show that (1, 1, 1) is the minimum on [0,3]×[0,3]×[0,3].

The function f is continuous and [0,3]×[0,3]×[0,3] is closed.

Therefore f subject to g=0 must have a minimum at the boundary or

at a critical point.

Minimum cannot be at the left edge, e.g., x0 = 0, because that would

fail to satisfy the constraint on g.

And minimum cannot be at the right edge, e.g., x0 = 3, because (1) would

force one of y0, z0 to be <= 0.

Therefore the minimum must be at a critical point, and (1, 1, 1) is the only one.

(b)

Let

u = (xy)²,

v = (yz)²,

w = (zx)²

Under the constraint x>=0, y>=0, z>=0

the condition uvw = 1 is equivalent to xyz = 1 for the following reason.

uvw = 1 iff

(xy)²(yz)²(zx)² = 1 iff

(xyz)^4 = 1 iff

xyz = 1

From (a) we know that the minimum of u + v + w is 3 under the constraint uvw = 1.

Therefore 3 is also the minimum of u + v + w under the constraint xyz = 1.

Even after adding the constraints x>=0, y>=0, z>=0 the minimum is still 3, because

the minimum occur for u = v = w = 1, which can be satisfied even under the

constraint x>=0, y>=0, z>=0.

(c)

Let

A = (a,0,0),

B = (0,b,0),

C = (0,0,c)

The area of the triangle ABC equals |AB×AC|/2.

In a coordinate system where the origin is shifted to A

A = (0,0,0)

B = (a,b,0)

C = (a,0,c)

AB×AC = (a,b,0)×(a,0,c) = (bc, ac, -ab)

The area of the triangle squared is

|AB×AC|²/4 = ((ab)² + (ac)² + (bc)²)/4

From part (b), under the constraint abc = 1,

the minimum of (ab)² + (ac)² + (bc)² is 3.

Therefore the minimum area of the triangle squared is 3/4,

and the minimum area of the triangle is √3/2.