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UPSC 2014 Maths Optional Paper 1 Q3b — Step-by-Step Solution 20 marks · Section A
Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 14 yrs · Read the full method →
Question
Find the maximum or minimum values of x 2 + y 2 + z 2 x^{2}+y^{2}+z^{2} x 2 + y 2 + z 2 a x 2 + b y 2 + c z 2 = 1 ax^{2}+by^{2}+cz^{2}=1 a x 2 + b y 2 + c z 2 = 1 l x + m y + n z = 0 lx+my+nz=0 l x + m y + n z = 0
Technique
Two-constraint Lagrange multipliers; trick of multiplying by the variable to identify λ = r 2 \lambda=r^{2} λ = r 2
Solution
Strategy. Two constraints ⇒ Lagrangian with two multipliers λ , μ \lambda,\mu λ , μ
Step 1 — Lagrange equations
L = ( x 2 + y 2 + z 2 ) − λ ( a x 2 + b y 2 + c z 2 − 1 ) − μ ( l x + m y + n z ) . L=(x^{2}+y^{2}+z^{2})-\lambda(ax^{2}+by^{2}+cz^{2}-1)-\mu(lx+my+nz). L = ( x 2 + y 2 + z 2 ) − λ ( a x 2 + b y 2 + c z 2 − 1 ) − μ ( l x + m y + n z ) . Setting partial derivatives to zero:
2 x − 2 a λ x − μ l = 0 ⟹ x ( 1 − a λ ) = μ l 2 . ( ∗ ) 2x-2a\lambda x-\mu l=0\;\Longrightarrow\;x(1-a\lambda)=\frac{\mu l}{2}.\tag{$\ast$} 2 x − 2 aλ x − μ l = 0 ⟹ x ( 1 − aλ ) = 2 μ l . ( ∗ ) Similarly y ( 1 − b λ ) = μ m / 2 y(1-b\lambda)=\mu m/2 y ( 1 − bλ ) = μ m /2 z ( 1 − c λ ) = μ n / 2 z(1-c\lambda)=\mu n/2 z ( 1 − c λ ) = μ n /2
Step 2 — Identify λ = r 2 \lambda=r^{2} λ = r 2
Multiply each equation in (\ast) by the corresponding variable and sum:
∑ i x i 2 ( 1 − ( coef i ) λ ) = μ 2 ∑ i ( coef i ) x i . \sum_{i}x_i^{2}(1-(\text{coef}_i)\lambda)=\frac{\mu}{2}\sum_{i}(\text{coef}_i)x_i. i ∑ x i 2 ( 1 − ( coef i ) λ ) = 2 μ i ∑ ( coef i ) x i . Concretely:
( x 2 + y 2 + z 2 ) − λ ( a x 2 + b y 2 + c z 2 ) = μ 2 ( l x + m y + n z ) . (x^{2}+y^{2}+z^{2})-\lambda(ax^{2}+by^{2}+cz^{2})=\frac{\mu}{2}(lx+my+nz). ( x 2 + y 2 + z 2 ) − λ ( a x 2 + b y 2 + c z 2 ) = 2 μ ( l x + m y + n z ) . Substitute the constraints a x 2 + b y 2 + c z 2 = 1 ax^{2}+by^{2}+cz^{2}=1 a x 2 + b y 2 + c z 2 = 1 l x + m y + n z = 0 lx+my+nz=0 l x + m y + n z = 0
r 2 − λ = 0 ⟹ λ = r 2 , where r 2 = x 2 + y 2 + z 2 at the critical point . r^{2}-\lambda=0\;\Longrightarrow\;\boxed{\lambda=r^{2},\text{ where }r^{2}=x^{2}+y^{2}+z^{2}\text{ at the critical point}.} r 2 − λ = 0 ⟹ λ = r 2 , where r 2 = x 2 + y 2 + z 2 at the critical point . So the Lagrange multiplier equals the squared distance at the extremum.
Step 3 — Determinantal equation for r 2 r^{2} r 2
From (\ast), x = μ l 2 ( 1 − a r 2 ) x=\dfrac{\mu l}{2(1-ar^{2})} x = 2 ( 1 − a r 2 ) μ l 1 − a r 2 ≠ 0 1-ar^{2}\ne 0 1 − a r 2 = 0 l x + m y + n z = 0 lx+my+nz=0 l x + m y + n z = 0
μ l 2 2 ( 1 − a r 2 ) + μ m 2 2 ( 1 − b r 2 ) + μ n 2 2 ( 1 − c r 2 ) = 0. \frac{\mu l^{2}}{2(1-ar^{2})}+\frac{\mu m^{2}}{2(1-br^{2})}+\frac{\mu n^{2}}{2(1-cr^{2})}=0. 2 ( 1 − a r 2 ) μ l 2 + 2 ( 1 − b r 2 ) μ m 2 + 2 ( 1 − c r 2 ) μ n 2 = 0. If μ ≠ 0 \mu\ne 0 μ = 0
l 2 1 − a r 2 + m 2 1 − b r 2 + n 2 1 − c r 2 = 0. \boxed{\;\dfrac{l^{2}}{1-ar^{2}}+\dfrac{m^{2}}{1-br^{2}}+\dfrac{n^{2}}{1-cr^{2}}=0.\;} 1 − a r 2 l 2 + 1 − b r 2 m 2 + 1 − c r 2 n 2 = 0. This is the determinantal equation in r 2 r^{2} r 2
l 2 ( 1 − b r 2 ) ( 1 − c r 2 ) + m 2 ( 1 − a r 2 ) ( 1 − c r 2 ) + n 2 ( 1 − a r 2 ) ( 1 − b r 2 ) = 0 , l^{2}(1-br^{2})(1-cr^{2})+m^{2}(1-ar^{2})(1-cr^{2})+n^{2}(1-ar^{2})(1-br^{2})=0, l 2 ( 1 − b r 2 ) ( 1 − c r 2 ) + m 2 ( 1 − a r 2 ) ( 1 − c r 2 ) + n 2 ( 1 − a r 2 ) ( 1 − b r 2 ) = 0 , a quadratic in r 2 r^{2} r 2 r 1 2 r_1^{2} r 1 2 r 2 2 r_2^{2} r 2 2
Step 4 — Geometric interpretation
The constraint surface Σ : a x 2 + b y 2 + c z 2 = 1 \Sigma:\,ax^{2}+by^{2}+cz^{2}=1 Σ : a x 2 + b y 2 + c z 2 = 1 1 / a , 1 / b , 1 / c 1/\sqrt a,\,1/\sqrt b,\,1/\sqrt c 1/ a , 1/ b , 1/ c
The plane π : l x + m y + n z = 0 \pi:\,lx+my+nz=0 π : l x + m y + n z = 0 Σ \Sigma Σ central section of the ellipsoid by π \pi π
Geometric meaning of r 2 r^{2} r 2 r = x 2 + y 2 + z 2 r=\sqrt{x^{2}+y^{2}+z^{2}} r = x 2 + y 2 + z 2 r 1 , r 2 r_1, r_2 r 1 , r 2 semi-major and semi-minor axis lengths of the elliptical section.
Answer
r 1 and r 2 are the semi-axes of the elliptical section of the ellipsoid by the plane. \boxed{\;r_1\text{ and }r_2\text{ are the semi-axes of the elliptical section of the ellipsoid by the plane.}\;} r 1 and r 2 are the semi-axes of the elliptical section of the ellipsoid by the plane.