hpas 2014 MATHS p1

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Standard Result: Euler’s Theorem on Homogeneous Functions

Statement: If a function f(x, y) is a homogeneous function of degree n in x and y, then its second-order partial derivatives satisfy the identity:

Principle of Linearity: Since the differential operator L = x^2 \frac{\partial^2}{\partial x^2} + 2xy \frac{\partial^2}{\partial x \partial y} + y^2 \frac{\partial^2}{\partial y^2} is linear, for any combination u = v + w, L(u) = L(v) + L(w).

Step 1: Decompose the Function and Identify Homogeneity Degrees

We split the given function u into two distinct parts:

where v(x, y) = x \phi(y/x) and w(x, y) = \psi(y/x).

• Testing v(x, y): Let x \to tx and y \to ty.
  v(tx, ty) = (tx) \phi\left(\frac{ty}{tx}\right) = t^1 \cdot x\phi(y/x) = t^1 v(x, y)
  Thus, v(x, y) is a homogeneous function of degree n_1 = 1.


• Testing w(x, y): Let x \to tx and y \to ty.
  w(tx, ty) = \psi\left(\frac{ty}{tx}\right) = \psi(y/x) = t^0 w(x, y)
  Thus, w(x, y) is a homogeneous function of degree n_2 = 0.

Step 2: Apply Euler’s Theorem to Each Part

Using the second-order extension of Euler’s Theorem for both functions independently:

For v(x, y) where n_1 = 1:

For w(x, y) where n_2 = 0:

Step 3: Combine via Linearity

By substituting u = v + w into our primary partial differential equation operator:

Final Conclusion:

The total sum vanishes completely based entirely on properties of homogeneity:

Hence Proved.

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Standard Result: Method of Lagrange Multipliers

Statement: To maximize or minimize an objective function f(x,y,z) subject to a constraint g(x,y,z) = 0, we formulate the Lagrangian L(x,y,z,\lambda) = f(x,y,z) - \lambda g(x,y,z). The extrema occur at the critical points where the partial derivatives of L with respect to x, y, z are all equal to zero.

Step 1: Set up the Geometry and Objective Function

Let the sphere be centered at the origin with a fixed radius R. The equation of the sphere is:
x^2 + y^2 + z^2 = R^2

Let the rectangular solid be inscribed within this sphere, also centered at the origin. Let the vertex of the solid in the first octant be (x, y, z). Because of symmetry, the full side lengths of the solid along the x, y, and z axes will be 2x, 2y, and 2z respectively.

The objective function we want to maximize is the Volume V:
V(x,y,z) = (2x)(2y)(2z) = 8xyz

Our constraint function is that the vertices must lie exactly on the surface of the sphere:
g(x,y,z) = x^2 + y^2 + z^2 - R^2 = 0

Step 2: Construct the Lagrangian

Using the Lagrange multiplier \lambda, we create the auxiliary function L:
L(x,y,z,\lambda) = 8xyz - \lambda(x^2 + y^2 + z^2 - R^2)

Step 3: Calculate Partial Derivatives

To find the critical points for maximum volume, take the partial derivatives of L with respect to x, y, and z, and set them to zero:

• \frac{\partial L}{\partial x} = 8yz - 2\lambda x = 0 \implies 4yz = \lambda x
• \frac{\partial L}{\partial y} = 8xz - 2\lambda y = 0 \implies 4xz = \lambda y
• \frac{\partial L}{\partial z} = 8xy - 2\lambda z = 0 \implies 4xy = \lambda z

Step 4: Solve the System of Equations

To eliminate \lambda, we can isolate it in each of the three equations. Assuming x, y, z \neq 0 (since a zero length would mean zero volume, which is a boundary minimum):

• From the first equation: \lambda = \frac{4yz}{x}
• From the second equation: \lambda = \frac{4xz}{y}
• From the third equation: \lambda = \frac{4xy}{z}

Equating the first two expressions for \lambda:
\frac{4yz}{x} = \frac{4xz}{y} \implies y^2 = x^2

Equating the second and third expressions for \lambda:
\frac{4xz}{y} = \frac{4xy}{z} \implies z^2 = y^2

Therefore, we have established that x^2 = y^2 = z^2. Since lengths must be strictly positive, we conclude:
x = y = z = \frac{R}{\sqrt{3}}

Step 5: Prove the Critical Point is a Maximum

We must verify that x = y = z yields the maximum possible volume, not a minimum.

Method 1: The Extreme Value Theorem (Calculus Argument)
The variables x, y, z are constrained to the first octant of a sphere (x,y,z \ge 0 and x^2+y^2+z^2=R^2), which is a closed and bounded (compact) set. By the Extreme Value Theorem, the continuous function V = 8xyz must have an absolute maximum and minimum on this set.
On the boundary of this domain (where x=0, y=0, or z=0), the volume is V = 0 (the minimum). Since the volume is strictly positive (V > 0) at our sole interior critical point x = y = z = R/\sqrt{3}, this point must be the global maximum.

Method 2: AM-GM Inequality (Algebraic Proof)
For any non-negative real numbers x^2, y^2, z^2, the Arithmetic Mean is greater than or equal to the Geometric Mean (AM \ge GM):
\frac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{x^2 y^2 z^2}
Substitute the constraint x^2 + y^2 + z^2 = R^2:
\frac{R^2}{3} \ge (xyz)^{2/3} \implies \left(\frac{R^2}{3}\right)^{3/2} \ge xyz
Multiply by 8 to get the volume V:
V = 8xyz \le 8\left(\frac{R^2}{3}\right)^{3/2}
The AM-GM inequality states that equality (and thus the maximum value of V) holds if and only if the terms are equal, i.e., x^2 = y^2 = z^2 \implies x = y = z.

Final Conclusion:

Because x = y = z yields the guaranteed maximum volume, the side lengths of the rectangular solid (2x, 2y, 2z) are all exactly equal. A rectangular solid with all equal side lengths is, by definition, a cube.

Hence Proved.

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