UPSC 2020 Maths Optional Paper 1 Q4c — Step-by-Step Solution

20 marks · Section A

Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 13 yrs · Read the full method →

Question

Find an extreme value of the function $u=x^2+y^2+z^2$ , subject to the condition $2x+3y+5z=30$ , by using Lagrange’s method of undetermined multiplier.

Technique

Lagrange multipliers — solve $\nabla u=\lambda\nabla g$ for $(x,y,z)$ in terms of $\lambda$ , substitute into the constraint.

Solution

Step 1 — Set up the Lagrange conditions

Constraint $g(x,y,z)=2x+3y+5z-30=0$ . Form

L=x^2+y^2+z^2-\lambda(2x+3y+5z-30).

Stationary conditions $\nabla u=\lambda\nabla g$ :

\frac{\partial L}{\partial x}=2x-2\lambda=0,\quad \frac{\partial L}{\partial y}=2y-3\lambda=0,\quad \frac{\partial L}{\partial z}=2z-5\lambda=0.

Hence

x=\lambda,\qquad y=\frac{3\lambda}{2},\qquad z=\frac{5\lambda}{2}.

Step 2 — Apply the constraint

2(\lambda)+3\left(\frac{3\lambda}{2}\right)+5\left(\frac{5\lambda}{2}\right)=30,

2\lambda+\frac{9\lambda}{2}+\frac{25\lambda}{2}=30\ \Rightarrow\ 2\lambda+17\lambda=30\ \Rightarrow\ 19\lambda=30,

\lambda=\frac{30}{19}.

Step 3 — Critical point and extreme value

x=\frac{30}{19},\qquad y=\frac{45}{19},\qquad z=\frac{75}{19}.

u=x^2+y^2+z^2=\frac{30^2+45^2+75^2}{19^2}=\frac{900+2025+5625}{361}=\frac{8550}{361}=\frac{450}{19}.

Answer

\boxed{\;u_{\min}=\frac{450}{19}\approx23.68,\quad\text{at }\left(\tfrac{30}{19},\tfrac{45}{19},\tfrac{75}{19}\right).\;}