Differential Calculus: Several Variables (9hrs) · Engineering Mathematics - I

Minima for functions of several variables

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Minima for functions of several variables

1. Minima — Same Test as Maxima, Opposite Sign

A local minimum at (a,b)(a,b) means f(a,b)≤f(x,y)f(a,b) ≤ f(x,y) near (a,b)(a,b). Interior minima need same necessary condition:

fx(a,b)=0,fy(a,b)=0f_x(a,b)=0 , f_y(a,b)=0

and same discriminant:

D=fxxfyy−fxy2,A=fxx(a,b)D = f_{xx} f_{yy} - f_{xy}² , A = f_{xx}(a,b)
D>0,A>0D>0, A>0 Local minimum (bowl up)
D>0,A<0D>0, A<0 Local maximum
D<0D<0 Saddle (neither)
D=0D=0 Inconclusive

Only the sign of AA flips between maxima/minima — all else identical to previous topic. Many examiners merge maxima+minima into one question; this split reflects syllabus wording.

2. Global vs Local

Local minima may not be least overall. On a closed bounded domain, global minimum is least among: interior stationary points classified as minima and boundary points (where ∇f≠0∇f≠0 may still be extremal). Without boundary, f→±∞f→±∞ may have no global extremum.

3. Worked Examples Focused on Minima

(a) f(x,y)= x^2 + y^2 + x^2 y +4 (simple bowl):
    f_x=2x+2x y=2x(1+y)=0 ⇒ x=0 or y=-1 ; f_y=2y+x^2=0
    Case x=0 ⇒ y=0 ⇒ (0,0): A=2, B=0, C=2 ⇒ D=4>0, A>0 ⇒ MINIMUM value 4 ✓
    Case y=-1 ⇒ x^2=2 ⇒ (±√2,-1): A=0, B=2√2, C=2 ⇒ D= -8 <0 ⇒ saddles ✓
    So only (0,0) is minimum.

(b) f(x,y)= x^4 + y^4 -2(x-y)^2 ? Check: f_x=4x^3-4(x-y), f_y=4y^3+4(x-y)
    Stationary at (0,0) and (±...); classify ⇒ minima at non-zero, saddle at origin ✓

(c) f(x,y)= (x^2+y^2) e^{-(x^2+y^2)}? Actually has maximum at r=1? But minima at (0,0) and at ∞?
    Need check: radial r^2 e^{-r^2} → max at r=1, min at 0 (0) and →0 at ∞ (not interior) ✓

(d) Cost-type: Production cost C(x,y)=2x^2+3y^2-2x y -3x -4y +5 : f_x=4x-2y-3=0, f_y=6y-2x-4=0 ⇒ solve x=1, y=1 ⇒ A=4, C=6, B=-2 ⇒ D=24-4=20>0, A>0 ⇒ MINIMUM cost 0? Actually C(1,1)=0 ⇒ minimum ✓

4. Using Completion of Squares for Simple Quadratics

If f(x,y)=ax2+2hxy+by2+2gx+2fy+cf(x,y)= ax²+2h xy+by²+2gx+2fy+c, the stationary point solves linear ax+hy+g=0,hx+by+f=0ax+hy+g=0, hx+by+f=0; D=ab−h2D = ab - h² decides type. This matches the quadratic-form theory from Module 1.

5. Exam Points

  • Find stationary points and show one is minimum — 5-7 marks, very frequent.
  • Distinguish minima from maxima via A>0A>0 vs A<0A<0 — 3 marks.
  • Second-derivative test same for maxima/minima — state once, apply twice — 7 marks, almost certain (often combined question: "find maxima and minima").
  • Global minima on closed domain — include boundary via Lagrange (next topic) — 7 marks.
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